FAN 

AN ALLUVIAL FAN FLOODING 

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COMPUTER PROGRAM 

USER'S MANUAL AND PROGRAM DISK 

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September 1990 

Federal Emergency Management Agency 
Federal Insurance Administration 


,



FAN 


An Alluvial Fan Flooding 
Computer Program 


User's Manual 
and 
Program Disk 


Prepared by 


Risk Studies Division 
Office of Risk Assessment 
Federal Insurance Administration 
Federal Emergency Management Agency 


September 1990 



TABLE OF CONTENTS 


Page 

Section 1-Introduction 
1-1 

1.1 Background........................................... 1-1 


1.2 Manual Format 
1-1 

1.3 Understanding the Methodology 
1-2 

Section 2 -Derivation 
2-1 

2.1 Basic Approach 
2-1 

2.2 Avulsions 
2-7 

2.3 Correction of High Flow Values 
2-8 

2.4 Multiple Channels 
2-9 

Section 3 -FAN Program Description 
3-1 

3.1 Program Purpose 
3-1 

3.2 Program Structure 
3-2 

3.3 Program Flow 
3-2 

3.4 Hardware and Software Requirements 
3-6 

3.4.1 Installation............................................ 3-6 


3.4.2 Running the Program 
3-6 

Section 4 -FAN Program Input and Output 
4-1 

4.1 Input Requirements 
4-1 

4.1.1 
Options for Entering Flood-Frequency Data 4-1 

4.1.1.1 
Option 1 -Entering Statistics of the Distribution 4-1 

4.1.1.2 
Option 2 -Entering Pairs of Recurrence Intervals 
and Discharge Values 4-1 

4.1.2 
Avulsion Factor 4-2 

4.1.3 
Data Requirements for the Multiple-Channel Option 4-2 

4.2 
Output 4-2 

4.2.1 
Flood-Frequency Data and Avulsion Factor. . . . . . . . . . . . . . . . 4-2 

4.2.2 
Single-Channel Region Mapping Parameters 4-3 

4.2.3 
Multiple-Channel Region Mapping Parameters. . . . . . . . . . . . 4-3 

Section 5 -FAN Program Example Runs................. 
5-1 


5.1 
Introduction........................................... 5-1 


5.2 
Example 1-Flood-Frequency Curve Defined by Mean 
Standard Deviation, and Skew Coefficient 5-4 

5.3 
Example 2 -Flood-Frequency Curve Defined by Pairs 
of Recurrence Intervals and Discharges 5-12 

5.4 
Output from Examples 5-28 

5.5 
Delineating Flood Insurance Zone Boundaries 5-37 

Section 6 -References 
6-1 


TABLE OF CONTENTS (Cont'd) 

Appendix A -Supplement to Derivation 

A.1 Hydraulic Characteristics: Single-Channel Region 
A.2 Hydraulic Characteristics: Multiple-Channel Region 
A.3 Derivation of Equation (2.11) 
A.4 Derivation of Equation (2.18) 
Appendix B-Listing 

Attachment-FAN Program Diskette (5-1/4 inch) 

LIST OF ILLUSTRATIONS 

Figures 

3-1 Flow of FAN.BAT . 
3-2 Flow of FANRUN.BAS Program . 
3-3 Flow of Contour Width Computations Corresponding to 

Selected Energy Depth . 
3-4 Flow of Contour Width Computations Corresponding to 

Selected Velocity . 
5-1 Example Alluvial Fan , . 
5-2 Example Flood-Frequency Curve . 
5-3 Page 1 of Output for Example Number 1 . 
5-4 Page 2 of Output for Example Number 1 . 
5-5 Page 1 of Output for Example Number 2 . 
5-6 Page 2 of Output for Example Number 2 . 
5-7 Page 3 of Output for Example Number 2 . 
5-8 Alluvial Fan Boundaries . 
5-9 "Smooth" Contours for Width Measurements . 
5-10 Depth Zone Boundaries (Single Channel Region) . 
5-11 Flood Insurance Zones (Single Channel Region) . 
5-12 Flood Insurance Zones (Multiple Channels Region) . 

LIST OF TABLES 

Table 

5-1 Probabilities and Corresponding Values of k (Skew =0) .... 

Page 

A-1 
A-1 
A-3 
A-5 
A-6 

B-1 

3-2 

3-3 

3-4 

3-5 
5-2 
5-3 
5-29 
5-30 
5-31 
5-32 
5-33 
5-38 
5-39 
5-41 
5-42 
5-43 

ii 


SECTION 1 -INTRODUCTION 


BACKGROUND AND PURPOSE 

In 1968, the National Flood Insurance Program (NFIP) was created as a multifaceted 
program for addressing the rising costs of flood damage. Since the passage of the 
Flood Disaster Protection Act of 1973, a major emphasis of the NFIP, which is 
administered by the Federal Emergency Management Agency (FEMA), has been 
identifying and mapping flood hazards in flood-prone communities nationwide. 
This identification and mapping effort has resulted in the evaluations of flood risks 
by detailed methods in more than 11,000 communities and evaluations by 
approximate methods for an additional 7,000 communities. 

The flood risk data that were developed from these evaluations have been published 
on Flood Insurance Rate Maps (FIRMs) and in Flood Insurance Study (FIS) reports. 
These data provide the basis for flood insurance premium rates as well as for local 
floodplain management measures required for participation in the NFIP. 

By 1979, FEMA recognized that standard procedures for evaluating riverine flood 
risks could not be used to evaluate flood risks attendant to alluvial fan flooding. 
Alluvial fan flooding is flooding that occurs on the surface of an alluvial fan or 
similar landform which originates at the "apex" and is characterized by high-velocity 
flows; active processes of erosion, sediment transport, and deposition; and 
unpredictable flow paths. Apex means a point on an alluvial fan or similar landform 
below which the flow path of the major stream that formed the fan becomes 
unpredictable and alluvial fan flooding can occur. 

Some of the flood hazards associated with alluvial fan flooding are flash flooding, 
unpredictable flow paths, and a high velocity of flow coupled with the material of 
the landforms being highly susceptible to erosion. FEMA recognized these 
significant floods hazards and the high level of interest in development on alluvial 
fans, and adopted, for flood insurance purposes, a methodology for evaluating and 
mapping flood hazards on alluvial fans. 

The computer program, FAN, discussed in this manual, was developed by FEMA to 
assist users in applying this methodology. 

MANUAL FORMAT 

This manual is comprised of six sections and two appendixes. The topics covered are 
described below. 

• Section 1 
Introduction 
• 
Section 2 Derivation of the method for mapping special flood hazards 
on alluvial fans 
• Section 3 
Description ofthe FAN Program 
• Section 4 
Input requirements and output 
• Section 5 
FAN Program example runs 
• Section 6 
References 
• Appendix A Supplement to Derivation 
• Appendix B Listings 
1-1 


UNDERSTANDING THE METHODOLOGY 

The FEMA methodology for determining flood hazards from alluvial fan flooding is 
simply the application of the definition for the 100-year flood. The application of 
the definition that is used in the FAN program is discussed in Section 2. The 
following analogy is presented to familiarize program users with such an 
application. 

A small structure is to be built at a point on the surface of and 477 feet from, the 

center of a large cone. However, a man who lives at the peak of the cone makes 

building this structure a problem. The man has a collection of iron balls, ranging in 

diameter from 10 to 60 feet. Once a year, this man rolls a die and, depending on the 

outcome of the roll, chooses a ball from his collection. If he rolls a "1" he chooses a 

10-foot-diameter ball; if he rolls a "2" he chooses a 20-foot-diameter ball; and so on. 

The man takes the chosen ball, places it at the peak of the cone, and lets it go. The 

ball rolls down the cone, taking an unpredictable path, and flattens anything in its 

way. 

The structure could be built so that it will withstand the impact of an iron ball-but 
,it is not practical to construct it to withstand the impact of a ball with a diameter of 
more than 20 feet. Therefore, the risk in any given year of being hit by a ball that 
has a diameter of more than 20 feet must be calculated. Two uncertainties are to be 

considered in this calculation: 

1. The outcome of the roll of the die is unknown. 
2. The path that the ball will follow is unknown. 
Because the probability of a ball taking anyone path is the same as the probability 
of it taking any other path, the probability of our structure being hit by the ball is 
equal to the diameter of the ball divided by the circumference of the cone at the 
point on which the structure is to be built. At the building site, the circumference of 
the cone is 3,000 feet. Thus, if the man rolled a "2," then the probability of the 20foot-
diameter ball hitting the structure is 20 + 3,000 or 0.0067. The probability of a 
"2" being rolled is 0.1666. 

To account for both uncertainties, the definition of conditional probability is 
applied. If peA) is the probability that event A will occur and PCB) is the probability 
that event B will occur, then the probability that event A will occur, given that event 
B has occurred, is P(AB) + PCB) and is written P(AIB). P(AB) is the event of both A 
and Boccurring together. 

For the example, AB denotes the event that a certain number was rolled and the 
structure is hit by the corresponding ball. This is the event whose probability is to be 
computed. Thus, not knowing whether event B will occur, the probability that 
events A and Bwill occur can be calculated as the product P(AIB) PCB) =P(AB). The 
probability ofthe structure being hit by a 20-foot-diameter ba I can be calculated by 
multiplying the conditional probability given above by the probability that a "2" will 
be rolled. Therefore, the probability ofthe structure being hit by a 20-foot-diameter 
ball in any given year is 0.0067 + 6 =0.0011. 

Before calculating the probability of the structure being demolished, several events 
that would result in its destruction should be noted. Specifically, the structure will 
be destroyed if hit by a 30-, 40-, 50-, or 60-foot-diameter ball. Using the 
nomenclature established above, the probability of event C or event 0 occurring is 

1-2 



written P(CUD) =P(C) + P(D) -P(CD). In the example, C would denote the event 
that a certain number is rolled and the structure is hit by the ball (denoted AB 
above); D would denote the event that a different number is rolled and the 
structure is hit by the ball corresponding to that number. One roll of the die cannot 
result in two different numbers; therefore, when C includes the event of rolling one 
number and D includes the event of rolling another, the probability would be 
calculated as P(CD) =O. 

In that case, P(CUD) =P(C) + P(D). Therefore, the probability that the structure will 
be destroyed in any given year is the sum of the probabilities of it being hit by a 30-, 
40-,50-, or 60-foot-diameter ball: 

6 
P(destruction) = L Pk (destruction ID =10k) P (D =10k) 
k=3 

where Pk is the probability of our structure being hit by a ball of diameter D =10k 
feet. 

Thus, 

6 10k 1 

P(destruction) = L ( )(-) 

. k=3 2n(477) 6 

= 
10 L6 
k 

5724n k=3 

1805724n 

= 0.01 

A 30-foot-diameter ball could be called the 100-year ball. It is the ball with the 
diameter that is expected to be equaled or exceeded at the building site once in 100 
years, on the long-term average. Because the probability of destruction at a given 
point depends on the circumference of the cone at that point, other locations on the 
cone will have 100-year balls of different diameters. If regions of the cone defined 
by their respective 100-year balls are to be mapped, the probability of destruction 
must be set equal to 0.01 and the equation for the circumference of the cone for 
each size of ball must be solved. 

1-3 



For example, if the 2D-foot-diameter ball region is defined as that region bounded 
by the circles where the 1DD-year ball has a 20-or 3D-foot-diameter, then it is the 
surface of the frustum that is bounded by circles of radii, r, and is computed as 
follows: 

1 6 10k 1 

r =-L (-)(-) =530 feet 

0.01 k=2 2n 6 
1 6 10k1 

r =-L (-)(-) =477 feet 

0.01 k=3 2n 6 
Thus, the map of the cone will show concentric circles separating the different 
regions defined by their respective 1DO-year balls. 

If the man is replaced by Mother Nature, the die by a flood-frequency curve, the balls 
by the maximum peak discharge of the year, and the cone represents an area subject 
to alluvial fan flooding, a method for mapping special flood hazards on an alluvial 
fan can be derived. Tne approach is the same as that just described. However, the 
derivation is more complex and is presented in Section 2 and Appendix A. 

1-4 



SECTION 2 -DERIVATION 


The FEMA methodology for determining flood hazards from alluvial fan flooding is 
simply the application of the definition of a 100-year flood. The location of the flow 
path during an alluvial fan flooding event is unpredictable. To determine the 
probability of a given point on the fan surface being flooded as a result of a storm 
over the watershed, the probability of the storm occurring and the probability that 
the flowpath of the floodwaters including that point must be considered. Th is 
section presents the derivation that is the basis for how the FAN program computes 
the magnitude of 100-year flood hazards in areas subject to alluvial fan flooding. If 
users are to recognize assumptions made in the program that are not consistent with 
the particular field conditions that are being analyzed and make the appropriate 
adjustments in their analyses, they must understand the derivation presented in this 
section and in Appendix A. 

In the program, the assumptions made are: (1) The maximum peak discharge at the 
apex of the alluvial fan in any given year is independent of the peak discharge there 
in any other year; and (2) Those peak discharges are identically distributed from year 
to year. In short, the peak flows at the apex are independent and identically 
distributed. The flood paths at a given elevation (Le., on a given contour) are also 
assumed to be independent and identically distributed. Those assumptions lead to a 
definition of the 100-year flood discharge at a given point on the alluvial fan as the 
discharge that is expected to be exceeded at that point with a probability of 0.01 in 
any given year. 

To illustrate the use of that definition, a simple problem that is somewhat analogous 
to that of defining flood hazards from alluvial fan flooding was presented in 
Section 1.3. This section presents the derivation of a method for mapping special 
flood hazards on the alluvial fan. Appendix A presents a supplement to the 
derivation. 

BASIC APPROACH 

Let H be a random variable denoting the occurrence of flooding at a given point 
subject to alluvial fan flooding. That is, 

1 if the point is flooded 

H = 0 if the point is not flooded (2.1) 

Let Qbe a random variable denoting the peak discharge resulting from a storm over 
the watershed. If fQ is the probability density function (pdf) of Q, then the 
probability of the point being inundated by a flood with a peak discharge of at least 
<10 cubic feet per second (ds) is 

P(H=1)= fill PHIQ(l,q)fQ(q)dq 

(2.2)
tIo 

~here 
PH1Q(l,q) is the probability of the point being flooded, given that the peakdischarge IS q cfs. 

2-1 



The 100-year flood discharge at a given point is defined as that discharge, qLOO' for 
which the probability of the point being flooded by at least qlO cfs is 0.01 In any ,given 
year. That is, for each point subject to alluvial fan flooding, ~he 
100-year flood 
discharge is that qlOO which satisfies 

0.01 = J'" PH1Q(l,q)fQ(q)dq (2.3) 
QlOO 
Therefore, if the probability of a given point being flooded by a given discharge 
varies with its location, then so does the magnitude of the 1OO-year flood discharge. 

If the 100-year flood discharge is to be quantified and the flood insurance zones are 
to be mapped, the functions in the integrand in Equation (2.2) must be defined. The 
FAN program defines the conditional probability, P Q (l,q), as the ratio of the 
channel width formed by q cfs, w(q), and the width of ~e 
area subject to alluvial fan 
flooding, W, at the elevation of the point of concern. That is, 

(2.4) 
The width, W, is called the contour width. 

When the contour width is much greater than the channel width [W> >w(q)], 
Equation (2.4) is equivalent to saying that each point on the contour has the same 
probability of being flooded. The definition of PH1Q(l,q) that is assumed in the FAN 
program depends on the function w(q) describing a relationship between channel 
width and peak discharge. This function can be derived from the assumptions 
outlined by Dawdy (Reference 1). 

Consider a constant discharge, q, that creates a rectangular channel and flows at 
critical depth. Also, assume that this flow erodes the sides of the channel, resulting 
in an increase in channel width. Because the discharge is constant, an increase in 
channel width, w, must be accompanied by a decrease in depth, d, if the energy of a 
unit volume of water is to remain minimum. If this erosion continues until the 
change in width per change in depth equals -200, then the channel shape satisfies 

dw 

-= -200 

dd (2.5) 

where d~ 
denotes differentiation with respect to depth. 

2-2 



Using Equation (2.5) to define channel shape, the width, w(q), of a rectangular 
channel carrying a discharge q at critical velocity can be written 

w(q) = 9.408 q~ 
(2.6) 

The derivations of this equation are presented in Appendix A. 

Thus, given that a storm will produce a peak discharge of q cfs, the program assumes 
that the probability of a point being inundated by ttiat flood is 

w(q) 
PH1Q(1,q) =


.w 

q~ 


= 9.408(
2.7)

W 

where W is the width of the area subject to alluvial fan flooding at the elevation of 
the contour on which the point lies -the contour width. 

The pdf of Q, fQ, must also be defined. The log-Pearson Type "' distribution is used to 
define flood frequency in the program. 

Let Y=~og10 
Q..Thus, th~ 
probability that a storm producing a peak discharge of at 
least qo In any gIven year IS 

a> Ak(y-m)k-1 
P(Q ) -A(y-m) d (2.8)

>qo = r(k) ey

JYo 

where A, k, and m are the three parameters of the Pearson Type "' distribution,rce) is the gamma function, and y is the base 10 logarithm of the discharge, q. Using 
y in the depth-discharge relationship assumed above yields 

w(y) = 9.408(l0)0.4y 

= 9.408eO.4yln 10 

= 9.408eO.92y 

(2.9) 
2-3 



Using Equation (2.2), the pdf defined by the assumptions, the 100-year flood ,discharge 
at any point can be defined. It is that discharge, qlOO' that has a base 
10 logarithm such that loglO qlOO =Y100 and 

eD eO.92y Ak (y_m)k-l 

0.01 = P(H=I) = 9.408 --e-A(y-m) dy 
(2.10)
fYloo W r(k) 

where W is the contour width at the elevation of the point of interest. Rearranging 
Equation (2.10) yields 

(A')k (y _ m)k-1 •

9.408C fa:l -.\. (y-m) d

0.01 =--(2.11 )
r(k)e y

W Yloo 

where 
92m Ak

eo.
C=--


(2.12)
(A-O.92)k 

and 

. 

A = A-0.92 (2.13) 

A complete version of the derivation from Equation (2.11) to Equation (2.12) is 
presented in Appendix A. 

Note that the integrand above is the Pearson Type III distribution assumed in 
Equation (2.8) for 10glOQ with a change in the scaling parameter from Ato A'= A


0.92. Ais referred to as the scaling parameter because changing it is equivalent to 
rescaling the random variable. For example, the integral in Equation (2.11) is the 
probability [defined by Equation (2.8)] that lO-O.92m/.\.' QAt.\.' exceeds qlOO at the apex. 
Also, note that the C in Equation (2.12) may be undefined for values of A between 
and including 0 and 0.92. 
When the skew of the flood-frequency curve is 0, fQ is log-normal. Therefore, 
instead of Equation (2.10), we have 

(Y-11 )2

-t 


eQ92y

eD e a 

(2.14)
0.01 = f 9.408 ---dy 
Yloo W oV2n 
where p is the mean of Y and CJ is the standard deviation. 

2-4 



Equation (2.14) can be rearranged to 

_~Y~ll' 
)2 

(2.15)
9.408 C Ie
0.01 = CD ----==--dy 
W YoV2n
lOO 

where 

C = e0.9'4J. + 0.420 
2 (2.16) 

and 

(2.17)
p' = p + 0.9202 

The derivation for Equation (2.1 5) is provided in Appendix A. 

Note that the integrand above is the normal distribution describing the pdf of 
laglo Q with a change in the mean from p to p.' =P. + 0.9202. Again, the change is 
equivalent to rescaling the random variable. That is, the integral in Equation (2.1 5) 
is the probability that 100.9202 Qexceeds q,oo at the apex. 

Thus, by defining the flood-frequency distribution and the boundaries within which 
all possible flood paths lie, a 100-year flood can be defined for any point subject to 
alluvial fan flooding. That flood is defined by FEMA in terms of velocity, v, and 
energy depth, D. Energy depth is the specific energy above the bottom of the 
rectangular channel. The discharge associated with an energy depth of D feet is 

(2.18)
q= 274.4 02.5 

Similarly, the discharge associated with a velocity, v, is 

q= 0.1289v5 (2.19) 

2-5 



Thus, to find the contour on which, in any given year, each point has a 0.01 
probability of being inundated by a flood whose energy depth exceeds 0.5 foot, 
Equation (2.11) [or Equation (2.15) if the skew is zero] should be solved for the 
contour width, W, with 

= loglO (48.5) (2.20) 

That is, 

9.408C 
WO.
5 = 0.01 P(Z>logl048.5) (2.21) 

where C is given by Equation (2.12) [or (2.16) if the skew is zero] and P(Z>logloq) 
represents tne integral in Equation (2.11) [or (2.15) if the skew is zero). The contour 
width corresponding to a 100-year energy depth of 1.5 feet is 

W1.5= 940.8 C P(Z> log 10 756) (2.22) 

The flood insurance zone for the area bounded by lines at the elevations where the 
contour widths are Wo.s and W1.S is labeled as "ZONE AO, DEPTH 1". All contour 
widths corresponding to energy depths of the form D =n + 0.5, where n is an 
integer, and satisfying the condition 

274.4 D 2.5 <Q
100 (2.23) 

are computed, and the flood insurance zones are labeled accordingly. The upper 
limit defined by Condition (2.23) is simply a reiteration of the fact that, in any given 
year, the probability of being hit by a flood with a depth greater than that created 
by the 100-year flood discharge at the apex, QlO~ 
is less than or equal to 0.01. 
Condition (2.23) is discussed further in Subsection l..j. 

2-6 



The flood insurance zones in areas subject to alluvial fan flooding are also labeled 

,-with the 100-year flood velocities. The velocity zone boundary widths are computed 
using a method similar to the one used to compute the depth zone boundary widths. 
For example, the width corresponding to a 100-year velocity of 3.5 fps is computed 
by first determining the discharge associated with that velocity 

q = (0.1289) (3.5)5=67.7 (2.24) 

and then computing the width 

W=940.8 C P(Z> loglO 67.7) 

(2.25) 
All contour widths corresponding to velocities of the form v =n + 0.5, where n is an 
integer, and satisfying the condition 

48.5 < 0.1289v 5<QlOO 
(2.26) 
are computed, and the flood insurance zones are labeled accordingly. Setting the 
lower limit at 48.5 cfs is equivalent to saying that floods with energy depths of less 
than 0.5 foot do not create the special flood hazards associated with A zones on 
FIRMs. 

2.2 AVULSIONS 
During a flood, the flow may abandon one path and follow a new one. That 
occurrence, termed an avulsion, can result from floodwater overtopping a channel 
bank and creating a new channel. The overtopping may be caused by the sudden 
deposition of sediment and/or debris or by the undercutting and subsequent failure 
of a channel bank. Because points below the avulsion may be in the path taken by 
the floodflow either before or after the avulsion occurs, their probability of being 
hit by the flood is greater than if the avulsion had not occurred. 

That increase in probability is accounted for by multiplying 1 plus the probability of 
an avulsion by the probability of being hit. Thus, if, during any flood, the probability 
of an avulsion is 0.5, the probability P(H = I), would be multiplied by the factor 1.5, 
the avulsion factor. Including the notion of an avulsion factor in our previous 
discussion yields a probability that a point will be hit by a flood of a magnitude 
greater than q cfs in any given year of 

9.408AC 

P(H= 1) = W P(Z>loglOq) (2.27) 

where A is the avulsion factor. 

2-7 



Accounting for the uncertainty of an avulsion by using a constant factor implies that 
the avulsion occurs during the peak of the flood and upfan of the point in question. 
Because we use the program to model the entire area subject to alluvial fan 
flooding, the latter implication is equivalent to saying that the avulsion occurs at the 
apex. 

CORRECTION FOR HIGH FLOW VALUES 

Without Condition (2.23), 

(2.23) 
contour widths that correspond to discharges greater than the 100-year flood 
discharge at the apex could be calculated. This is seen by considering the general 
form of Equation (2.21), including an avulsion factor 

(2.28)
W=940.8ACP(Z > loglOq) 

Note that when P(Z > loglOq) is greater than 0, W is greater than O. Also note that 
P(Z>loglOq) is greater than 0 for all q when the skew of the flood-frequency curve is 
zero; for all q > 10m when the skew is greater than zero; and for all q < 10m when 
the skew is less than zero, where m is the Pearson Type III parameter in Equation 
(2.8). 

Thus, for discharge values greater than the 100-year flood discharge at the apex, 
there are values for W that satisfy Equation (2.28). This implies that the probability 
of a point being hit by a flood of a magnitude greater than the SOO-year flood 
discharge at the apex is 0.01 in any given year. The root of this contradiction is seen 
by reviewing the definition of the conditional probability given by Equation (2.4): 

(2.4) 
For very large values of q and relatively small values of W, w (q) may be greater than 

W. In that case, the probability given by Equation (2.4) of a given point being hit by 
q ds, given that that discharge is realized at the apex, is 
w(q) 
PH1Q(1,q) = W > 1 (2.29) 

which is absurd. 

2-8 



To avoid small errors introduced by the contradiction in Equation (2.29), the 
calculation of the contour width must be adjusted. If qwdenotes the discharge that 
creates a channel as wide as the contour width, the probability given by Equation 

(2.11) can be corrected by replacing the upper limit of the integral with log.loqw and 
adding the probability that qw is exceeded at the apex. That is, find the contour 
width, W, such that 
(2.30) 
where q. is the discharge associated with the depth or velocity being investigated, 
P(Z > IOglOq) is the integral in Equation (2.11), and P(Q > qw} is given by Equation 
(2.8). 

Including an avulsion factor, A, gives the final expression of the problem to be 
solved. That is, for each q. associated with the depths and velocities described above, 
the contour width, W, thch satisfies the following equation is found: 

(2.31) 
The W given by Equation (2.31) and the W given by Equation (2.28) differ by a small 
amount -negligible in most cases. For avulsion factors greater than 1.0, a solution 
of Equation (2.31) is not the exact solution to the problem. However, the error 
introduced by using avulsion factors greater than 1.0 is much smaller than the 
difference between widths given by Equations (2.28) and (2.31). It is mentioned here 
only as a matter of detail. 

If the path taken by the floodflow is as wide as the alluvial fan at some elevation, 
then avulsions above that elevation are impossible. Thus, for widths between 9.408 

qw215 

and 18.816 qw2/S, the avulsion factor accounts for more risk than is present. 
Again, the more riSk is negligible. 

MULTIPLE CHANNELS 

On many alluvial fans, floods do not remain within a single channel from the apex 
down to the toe. Instead, a flood may be carried by a single channel to some point 
down the fan and then by several channels below that point. The point at which the 
single channel becomes multiple channels is referred to as the bifurcation point. 
Analyses of several well-documented alluvial fan flooding events indicate that 

2-9 



the cumulative width of the multiple channels is 3.8 times the width of the single 
channel above the bifurcation point (Reference 2). Therefore, in the 
multiple-channel region, we redefine the width-discharge relationship given by 
Equation (2.7) as 

w(q) = (3.S)(9.40S)q215 = 35.7504 q215 (2.32) 

Note that the multiple channels may be regarded as equivalent to a single channel 
with a width 3.8 times greater than the width of a single channel above the 
bifurcation point. 

In addition, it is assumed that the depth ~nd 
velocity of floodflows in the multiplechannel 
region can be estimated using Manning's equation with the friction slope 
set equal to the slope of the alluvial fan (Le., a normal depth approximation). Thus, 
the relationship between depth of water, d, and discharge, q, is defined as 

(2.33) 
where n is the roughness coefficient (Manning's "n") and s is the slope of the alluvial 
fan. 

The velocity-discharge relationship can be written 

(2.34) 
The velocity head can be computed from Equation (2.34): 

y2 1( 2 

-=-0.3033 n -0.6 s0.3 q0.24)
2g 2g. 

(2.35) 
Therefore, the energy depth, D, in the multiple-channel region is 

y2 

D=d+ 


2g 

2-10 


To find the discharges associated with the energy depths used to map the depth 
zone boundaries in the multiple-channel region, Equation (2.36) must be solved for 
D = n + 0.5, where n is an integer. Those discharges are bounded above by Q, O' 
Similarly, to determine the discharges associated with the velocities used to map tRe 
velocity zone boundaries in the multiple-channel region, Equation (2.34) must be 
solved for v = n + 0.5, where n is an integer. The discharge values given by 
Equation (2.34) are restricted to values less than Q100 and greater than the discharge 
that satisfies Equation (2.36) with D =0.5. The restrictions on the discharges are 
analogous to Conditions (2.23) and (2.26) for the single-channel region. 

The problem to be solved for the multiple-channel region is the same as that 
expressed by Equation (2.31) with the expression in brackets multiplied by 3.8. That 
is, for each qj associated with the depths and velocities just described, the 
contour width, W, that satisfies 

(2.37) 
must be found. The FAN program computes the contour widths, corresponding to 
the depths and velocities of interest, that satisfy Equation (2.31) for the singlechannel 
region [and Equation (2.37) for the multiple-channel region]. The values of 
cumulative distribution functions (CDFs), denoted P(.) in those equations, are 
determined by linear interpolation using the Pearson Type III tables published in 
Guidelines for Determining Flood Flow Frequency, Bulletin No. 17B (Reference 3). 
The tables in the program contain the number of standard deviations (k-values) that 
a given value of the random variable is from the mean. The tables in the program 
contain k-values corresponding to various values of the CDF for skew values varying 
by 0.1 and ranging from -4.1 to 4.1. 

2-11 



SECTION 3 -FAN PROGRAM DESCRIPTION 


PROGRAM PURPOSE 

The primary purpose of the FAN program is to comRute the contour widths 
corresponding to the flood insurance zone boundaries (for example, the width of 
the area subject to alluvial fan flooding where the 100-year flood depth is 1.5 feetthe 
boundary between the 1-and 2-foot-depth zones). That purpose is 
accomplished by (1) using Equations (2.18) and (2.36) to determine the discharge (q) 
value that corresponds to each energy depth (D) value for the single-and multiplechannel 
regions, respectively; (2) using Equations (2.19) and (2.34) to determine the 
q value that corresponds to each velocity (v) value for the single-and multiplechannel 
regions, respectively; and (3) using Equations (2.31) and (2.37) to determine 
the contour width (W) value for the single-and multiple-channel regions, 
respectively. 

Probabilities, denoted pee) in Equations (2.31) and (2.37), are computed using th.e 
Pearson Type III tables published in Guidelines for Determinin~ 
Flood Flow 
Frequency (Reference 2). To use those tables, the program must deIne the floodfrequency 
curve in terms of its mean (11), standard deviation (a), and skew coefficient 
(G). The tables consist of pairs of probabilities and numbers (k-values) of standard 
deviations between the corresponding value and the mean of the random variable 
for each skew value. 

Given the statistics of the Pearson Type "' distribution, the discharge, qp' that has a 
probability of being exceeded in any given year is given by 

(3.1 )
loglO qp =0 k + ~ 


Similarly, given a discharge, q, and the Pearson Type III statistics, the probability is 
determined by computing k from Equation (3.1) and finding the corresponding 
probability through linear interpolation. 

Because the contour width depends on the probability that the rescaled 
(transformed) random variable exceeds the discharge in question [see 
Equation (2.11)], the mean, standard deviation, and skew coefficient of the rescaled 
random variable must be calculated to use the tables. For random variables that are 
Pearson Type III distributed, those statistics are related to the parameters of the 
distribution by 

k 
~=m+ 
(
3.2)

11 

Vk 

0=-(3.3)

11 

3-1 



2 


G=~(3.4)

vk 

Note that the skew coefficient, G, is independent of the scaling parameter, A, and, 
therefore, is the same for both distributions. The mean and standard deviation of 
the rescaled random variable are found by replacing Awith A' = A-0.92 in Equations 

(3.2) and (3.3), respectively. 
3.2 PROGRAM STRUCTURE 
Program flow is controlled by a DOS batch file, named FAN.BAT, that calls four 
programs written in BASIC (see Appendix B for listings). Compiled versions of those 
programs, PEARSN.EXE, FANINP.EXE, FANRUN.EXE, and AGAIN.EXE, reside in a 
directory named AL-FAN on the diskette that accompanies this manual. The order in 
which FAN calls those programs is shown in Figure 3-1. 

t-----------J·~I---PE-A-R-S-N-.E-X-E-


BRUN4S.EXE 

t .. 

FANINP.EXE 

,~ 


FANRUN.EXE 

, 

AGAIN.EXE 

Figure 3-1. Flow of FAN.BAT 

PEARSN.EXE assigns the k-values of the Pearson Type III tables. FANINP.EXE accepts 
the input data and defines the flood-frequency curve. These programs interact 
through BRUN4S.EXE, a copyrighted program of Microsoft Corporation, 1982-88. 
After the flood-frequency curve has been defined, FANRUN.EXE computes the flood 
risk data. AGAIN.EXE, a short program, gives users the option to make another run. 

3.3 PROGRAM FLOW 
Figure 3-2 shows the flow of the FANRUN.BAS program. The flow of the contour 
width computations corresponding to the selected energy depth and selected 
velocity is shown in Figure 3-3 and 3-4, respectively. The numbers shown in each box 
are the line numbers of FANRUN.BAS that perform the given task. A listing of the 
program is given in Appendix B. 

3-2 


Read Input Data 
(Lines 126-158) 

Transform Random Variable 
(Lines 2000-2028) 

, 
Compute Contour Widths 
For Depth Zone Boundaries 
(Lines 2200-2266) 

It 

Compute Contour Widths 
For Velocity Zone Boundaries 
(Lines 2400-2478) 


, 
Print Output Data 

Figure 3-2. Flow of FANRUN.BAS Program 

3-3 


Compute Discharge Corresponding to 
Selected Energy Depth 


(Line 2318) 


lit 

Compute Probability of Exceeding 
Rescaled Discharge at Apex 
(Lines 4000-4032) 
, 
Compute Contour Width 
(Lines 2238-2246) 
It 
Adjust Contour Width 
for Large Flow Values 
(Lines 4200-4234) 

Figure 3-.3. Flow of Contour Width Computations Corresponding to 
Selected Energy Depth 


3-4 



Compute Discharge Corresponding 
to Selected Velocity 
(Line 2432) 


Compute Probability of Exceeding 
Rescaled Discharge at Apex 
(Lines 4000-4032) 


,It 

Compute Contour Width 
(Lines 2450-2458) 

~ 


Adjust Contour Width 
for Large Flow Values 
(Lines 4200-4234) 

Figure 3-4. Flow of Contour Width Computations Corresponding 
to Selected Velocity 

3-5 


3.4 HARDWARE AND SOFTWARE REQUIREMENTS 
The FAN program can be run on any IBM personal computer (PC) or any IBMcompatible 
PC that uses DOS release 2.0 or higher. The program, which is supplied 
on the 5-1/4-inch diskette that accompanies this manual, is compatible with fixedand 
floppy-disk systems. This version of the program was written for single-user 
systems. 

3.4.1 INSTALLATION 
The program can be installed on another disk (hard or floppy) by inserting the 
enclosed diskette in Drive A and typing 

A:INSTALL <drive>: 

and pressing IENTER I· The notation <drive> represents the letter of the drive. If 

Drive C is the hard disk in the system, the program can be installed on the hard disk 

by typing 

A:INSTALL C: 

If a properly formatted floppy diskette is inserted in Drive B, the program can be 
installed by typing 

A:INSTALL B: IENTER I 

which copies the program onto that diskette. 

The installation process copies the batch file FAN.BAT to the root directory of the 
disk in the specified drive and creates a directory, AL-FAN. The executable files 
associated with the program are copied into that new directory. 

3.4.2 RUNNING THE PROGRAM 
The program is run by typing FAN in response to the prompt for the root directory of 
the correct drive. Ifthe prompt is for the correct drive but not the root directory, the 
user can access the root directory by entering CD\. 

For example, if the user inserts the diskette that accompanies this manual in Drive A 
and enters A, one of the following will appear on the screen: 

A:\> or A:\AL-FAN> 

3-6 


The first prompt, A:\>, is for the root directory. If A:\> appears, entering FAN in 
response to that prompt will allow the user run the program. If A:\AL-FAN> 
,-appears, entering CO\ accesses the root directory. 

The user is guided by prompts on the screen. The examples provided in Section 5 
illustrate the interaction between the user and the program. 

3-7 



SECTION 4 -FAN PROGRAM INPUT AND OUTPUT 

4.1 INPUT REQUIREMENTS 
The input data required to run the FAN program are entered as responses to 
prompts. Data should be entered one value at a time. Values are entered by 
pressing ENTER. Responses to yes or no questions may be entered by pressing Y or N 
(capital or lower case), and then pressing ENTER. 

The minimum input data required are flood-frequency data and the avulsion factor. 
If the user chooses to compute contour widths for the multiple-channel region, the 
alluvial fan slope and the roughness coefficient (Manning's "n") are also required. 

4.1.1 Options for Entering Flood-Frequency Data 
The program offers two options for entering flood-frequency data. Once the floodfrequency 
curve is defined, the program computes the lOa-year flood discharge. If 
the 100-year flood discharge value is less than 50 (cfs) or greater than 500,000 cfs, 
the program will not run. If the lOa-year flood discharge value is within that range, 
the program computes the product of the standard deviation and skew coefficient 
to see if the transformation constant, C, can be defined. [See Equation (2.12)] If that 
product is greater than 2.1, the program will not run. 

4.1.1.1 Option 1-Entering Statistics of the Distribution 
Option 1 for defining the flood-frequency curve is to enter the mean, standard 
deviation, and skew coefficient. Standard deviations cannot be less than 0.1 and 
skew coefficients must be within the range of -4.1 to 4.1. 

4.1.1.2 Option 2-Entering Pairs of Recurrence Intervals and Discharge Values 
Option 2 for defining the flood-frequency curve is to enter a minimum of three pairs 
of recurrence intervals and discharge values. The program finds the "best fit" of 
those data to a log-Pearson Type III distribution. The best fit is the log-Pearson 
Type III distribution that results in the maximum correlation coefficient of a leastsquares 
fit of the data. Restrictions on the input data are as follows: 

• 
Recurrenceintervalsmustbe between1.001 and1,000years. 
• 
Each recurrence interval may be entered only once (e.g., two la-year flood 
discharge values cannot be entered). 
• 
Discharge values must be greater than zero. 
• 
Discharge values cannot decrease as the recurrence interval increases (e.g., 
if the 1DO-year flood discharge value is 1,000 cfs, the gO-year flood 
discharge value cannot be 2,000 cfs). 
The least-squares fit is the straight line through a set of data pairs that minimizes the 
.sum of th~ 
squares of the differences between the given values in the range of the 
data and the corresponding values predicted by the straight line. For each skew 
value tested, the program computes the k-values that correspond to the entered 
recurrence intervals and the base 10 logarithms of the entered discharges. The 

4-1 



resulting pairs of data are then fit to the line defined by Equation (3.1), and the 
correlation coefficient is computed. The skew value that results in the greate~t 
correlation coefficient is the skew of the flood-frequency curve. That skew value IS 
found by treating the correlation coefficient as a function of skew with, at most, one 
critical point-its maximum. The mean and standard deviation of the floodfrequency 
curve are the slope and intercept, respectively, of the least-squares fit line 
for the chosen skew value. 

4.1.2 Avulsion Factor 
The avulsion factor may take any value. However, because the contour widths are 
proportional to the avulsion factor, using an avulsion factor of 0 will result in all 
contour widths being O. Therefore, if the user enters a value of 0, the program will 
change it to 1.0. 

4.1.3 Data Requirements for the Multiple-Channel Option 
If the user chooses to compute contour widths for the multiple-channel region, 
additional data are required. Specifically, the user must supply the alluvial fan slope 
value and the roughness coefficients to be used in Manning's equation. The alluvial 
fan slope values (dimensionless) are restricted to a range of 0.000001 to 1.0; 
roughness coefficients are restricted to a range of 0.001 to 1.0. If the user enters a 
value outside those ranges, a message on the screen will advise the user that the 
input is too small or too large and instruct the user to re-enter the values. 

4.2 OUTPUT 
Output from a run consists of two or three pages, depending on the options chosen 
by the user. The output is written to a file named FAN.OUT. When a run is 
completed, the user is given three options; (1) to view the output on the screen, (2) 
to print the output, and (3) to make another run. When a new run is executed the 
file FAN.QUT is erased to make room for the new output. Thus, if the user wishes to 
save the output from a previous run, the option to print the output must be selected. 

4.2.1 Flood-Frequency Data and Avulsion Factor 
The first page of output data lists the avulsion factor and information pertaining to 
the flood-frequency curve. The flood-frequency data consists of the following: 

• 
The option chosen to define the flood-frequency curve (If the option to 
enter pairs of flood-frequency data was selected, those data and the 
discharges corresponding to the entered recurrence intervals and defined 
by the least-squares fit of the data are listed.) 
• 
The mean, standard deviation, and skew ofthe flood-frequency curve 
• 
The 10-, SO-, 100-, and SOO-year flood discharges (for use in the "Summary 
of Discharges" table in the FIS report) 
• 
The scale change in (Le., transformation of) the random variable denoting 
log\O Q, the statistics of the distribution of the changed random variable, 
ana the transformation constant (e) 
4-2 


4.2.2 Single-Channel Region Mapping Parameters 
The second page of the output data lists the special flood hazard information for the 
flood insurance zone boundaries in the single-channel region. The information for 
the depth zones is given first; the information for the velocity zones is given second. 
That information is in tabular form and consists of the following: 

• 1OO-year energy depths and velocities 
• Depth of water associated with each energy depth or velocity 
• Discharge associated with each energy depth or velocity 
• Probability that that discharge is exceeded at the apex 
• Probability that the "rescaled discharge" is exceeded at the apex 
• Contour width associated with each energy depth or velocity 
4.2.3 Multiple-Channel Region Mapping Parameters 
The third page of the output data lists the special flood hazard information for the 
multiple-channel region, and consists of the following: 

• Slope of the alluvial fan 
• Roughness coefficient used in the energy depth and velocity computations 
• 100-year energy depths and velocities 
• Depth of water associated with each energy depth or velocity 
• Discharge associated with each energy depth or velocity 
• Probability that that discharge is exceeded at the apex 
• Probability that the "rescaled discharge" is exceeded at the apex 
• Contour width associated with each energy depth or velocity 
4-3 



SECTION 5 -FAN PROGRAM EXAMPLE RUNS 


INTRODUCTION 

The example runs in this section illustrate the interaction between the user and the 
FAN program. The alluvial fan used in the examples is shown in Figure 5-1. The 
flood-frequency curve for these examples is log-normal with a standard deviation of 

1.0 and a 2-year flood discharge of 10 cfs. Therefore, the distribution has a mean of 
1.0 and a skew of o. The flood-frequency curve is shown in Figure 5-2. The slope of 
the alluvial fan is 0.085 and the roughness coefficient is 0.05. The avulsion factor is 
1.0. 

Two examples have been selected to demonstrate the various options. In Example 
Number 1 (Subsection 5.2), the mean, standard deviation, and skew coefficient are 
used to define the flood-fre.quency curve; the option of computing contour widths 
for the multiple-channel region is not chosen. In Example Number 2 (Subsection 
5.3), flood-frequency data are entered in the form of pairs of recurrence interval and 
discharge values. The option of computing contour widths for the multiple-channel 
region is chosen. In the examples, user-supplied information is denoted by bold 
print. 

5-1 



1" = 1000' 


Figure 5-1. Example Alluvial Fan 

5-2 


1000 

w 
a:


" 
"
<(I) 
~~ 
100 

(I)
o 


10 


.998 .99 .98.95.9 .8.7.6.5.4.3.2 ..1 .05 .02.01 .002 

PROBABILITY OF EXCEEDING DISCHARGE AT APEX 
Figure 5-2. Example Flood-Frequency Curve 
5-3 


1000 

100 

10 


5.2 
EXAMPLE 1 -FLOOD-FREQUENCY CURVE DEFINED BY MEAN, 
STANDARD DEVIATION, AND SKEW COEFFICIENT 
,


When the user types FAN, the batch file is called, the program begins to run, 
and the following message appears on the screen: 

ALLUVIAL FAN FLOODING COMPUTER PROGRAM 

In response to prompts, enter data one 
value at a time, then press enter. 

Answer yes-or-no questions with the 
corresponding letters (i.e., Y or N). 

PEARSON TYPE-III TABLES BEING LOADED••••••• 

While that message is on the screen, the program assigns the k-values of the 
Pearson Type III distributions. That process takes a few seconds to complete. 
When the k-values have been assigned, the message changes. 

ALLUVIAL FAN FLOODING COMPUTER PROGRAM 

In response to prompts, enter data one 
value at a time, then press enter. 

Answer yes-or-no questions with the 
corresponding letters (i.e., Y or N). 

PRESS ENTER TO PROCEED.................. IENTER I 


5-4 



If the user presses IEN'kERa' ' the screen is cleared and the first promptappears. The user is as e to enter the name of the alluvial fan. Therefore, 
Example Number 1 is entered. 

P..... F1 alld thell p...u ENTER to ezit 

ENTER THE NAME OF THE ALLUVIAL FAN 

EXAMPLE NUMBER 1 

IENTER I 

The next prompt concerns the multiple-channel option. In the first example, 
contour widths will be computed for the single-channel region only. 
Therefore, the answer to the prompt is "N." 

Press F1 alld thell press ENTER to ezit 

DO YOU WISH TO COMPUTE ZONE BOUNDARIES 

FOR MULTIPLE CHANNELS (YIN)? N IENTER I 

5-5 



Next, the user is asked to enter the avulsion factor. An avulsion factor of 1.0 is 
entered. 

PI"OSS Fl alld thell press ENTER to e~t 


ENTER AVULSION FACTOR 1 IENTER I 

The next prompt concerns the option for defining the flood-frequency curve. 
The flood-frequency curve is to be defined by statistics; therefore, Option 1 is 
selected. 

PI"OS. Fl alld thell press ENTER to e~t 


YOU MAY DEFINE THE FLOOD FREQUENCY CUR""!: BY: 
(ll.... ENTERING THE MEAN. STANDARD DEVIATION, AND SKEW COEFnCIENT 


OF THE PEARSON TYPE·m DISTRIBUTION 
(2).... ENTERING (AT LEAST THREE) PAIRS OF RETURN INTERVALS AND DISCHARGES 
PLEASE ENTER OPTION NUMBER (lOR 2) 1 IENTER I 

5-6 



After choosing Option 1, three statistics are to be entered, one value at a time, 
beginning with the mean. 

Press Fl and than press ENTER to e:<it 

ENTER MEAN 1 

IENTER I 

The mean remains on the screen, and the user is prompted to enter the 
standard deviation. 

Press Fl and then press ENTER ~ 
.:<it 

ENTER II1EA.."1 1 

ENTER STANDARD DEVIATION 1 IENTER I 

5-7 



10..oo 
FLOOD FREQUENCY CURVE 
MEAN = 1 
STDDEV = 1 
SKEW =0 
10 

The mean and standard deviation remain on the screen, and the user is 
prompted to enter the skew coefficient. 

PrlIu F1 aDd tho.. prou ENTER to ozit 

ENTER MEAN 1 
ENTER STANDARD DEVIATION 1 


ENTER SKEW COEFFICIENT 0 IENTER I 

Entering the skew coefficient completes the data entry requirements. The 
screen is cleared and the flood-frequency curve is drawn. 

.999 .99 .9 .5 .1 .01 .001 

5-8 



When the computations of the contour widths are complete, the message 
"PRESS ENTER TO CONTINUE" appears atthe bottom of the screen. 

FLOOD FREQUENCY CURVE 

.999 .99 .9 .5 .1 .01 .001 
........ PRESS ENTER TO CONTINUE 


•..·····1 ENTER I 
10..oo10 I.....lo--lo__......._.-...........Iooo-J 
Pressing IrN~R 
I,clears the screen, and a picture of an alluvial fan 
(with simp e oundaries) is drawn. The depth (solid lines) and velocity 
(dashed lines) zone boundaries with the computed contour widths for the 
single-channel region are drawn on the alluvial fan. When the picture is 
complete, the "PRESS ENTER TO CONTINUE" message appears again. 

EXAMPLE NUMBER 1 

~3.5\ 


SINGLE·CHANNEL REGION 

.4.5 


\ 

\ 

~ 


" ,, 
___ 5).5\ \ \ 
6.~ 
\ ~~ 


<L" II'II I 

II, I 
,, I 

,, I 

1.5 
I ,! , 
I 
I,,, 
,: 
___DEPI'H 

---VELOCITY 
........ PRESS ENTER TO CONTINUE •••••••• 1ENTER I 

5-9 



When IENTERIis pressed, aprompt asks the userifhe/shewould liketo 
view the output. Instead of viewing the output, the user chooses to print it 
out. Therefore an "N" is entered. 

DO YOU WISH TO VIEW SOME OUTPUT (YINl? N IENTER I 

hi response to the next prompt, "y" is entered to activate the printing option. 

DO YOU WISH TO PRINT THE OUTPUT IY·'Nl' Y IENTER I 

As a result of this response, the output file is sent to the printer. A final 
prompt gives the user the option to make another run. 

5-10 


DO YOU WISH TO MAKE ANOTHER RUN (YIN)? YIENTER I 

By responding "Yes" to the prompt, the program returns the user to the 
beginning. 

5-11 



5.3 EXAMPLE 2 -FLOOD-FREQUENCY CURVE DEFINED BY PAIRS OF 
RECURRENCE INTERVALS AND DISCHARGES 
ALLUVIAL FAN FLOODING COMPUTER PROGRAM 

In response to prompts, enter data one 
value at a time, then press enter. 

Answer yes-or-no questions with the 
corresponding letters (i.e., Y or N). 

PEARSON TYPE-III TABLES BEING LOADED••••••• 

Again, the program assigns the k-values of the Pearson Type III distributions, 
and the message changes. . 

ALLUVIAL FAN FLOODING COMPUTER PROGRAM 

In response to prompts, enter data one 
value at a time, then press enter. 

Answer yes-or-no questions with the 
corresponding letters (i.e., Y or N). 

PRESS ENTER TO PROCEED•••••••••••••••••• 

IENTER I 

5-12 



The same procedure is followed as for the first example run. Therefore, the 
first entry is the alluvial fan name. 

P..... Fl aDd thlll p..... ENTER to nit 

ENTER THE NAME OF THE ALLUVIAL FAN 

EXAMPLE NUMBER 2 IENTER I 

For this example, the multiple-channel option is to be used. Therefore, the 
answer to the next prompt is "Yes." 

P",•• Fl alld thlll p,..•• ENTER to l:rit 

DO YOU WISH TO COMPUTE ZONE BOUNDARIES 

FOR MULTIPLE CHANNELS (YIN)? YIENTER I 

5-13 



Having selected the multiple-channel option, the user is first asked to enter 
the slope of the alluvial fan. 

P..u Fl IlIId the" p..u ENTER to nit 

ENTER SLOPE OF ALLUVIAL FAN .085 IENTER I 

The slope is displayed, and the user is prompted to enter the n-value. 

P..ss Fl and then p..ss ENTER tD .zit 

SLOPE = .085 

ENTER ROUGHNESS COEFFICIENT (N·VALUE) .05 IENTER I 

5-14 



The slope and n-value are displayed, and the user is asked to enter the 
avulsion factor. 

Pre.. Fl &lid then p.... ENTER to nit 

MULTIPLE CHANNEL PARAMETERS: 
SLOPE = .085 
N· VALUE = .05 

ENTER AVULSION FACTOR 1 

IENTER I 

The user now chooses the option for determining the flood-frequency curve. 
For this example, Option 2, least-squares fit of data, is selected. 

Press Fl and then pre.. ENTER to eZIt 

YOU MAY DEFINE THE FLOOD FREQUENCY CURVE BY: 

11).... ENTERING THE MEAN. STANDARD DEVIATION, AND SKEW COEFFICIENT 
OF THE PEARSON TYFE·m DISTRIBUTION 
12).... ENTERING (AT LEAST THREE) PAIRS OF RECURRENCE INTERVALS AND DISCHARGES 
PLEASE ENTER OPTION NUMBER (l OR2) 2 IENTER I 

5-15 



The first prompt in Option 2 asks for the number of pairs of data to be 
entered. 

,


Pre.. Fl aDd theD prell ENTER to uit 

HOW MANY PAIRS OF DISCHARGES AND 
RECURRENCE INTERVALS DO YOU WISH TO ENTER? 6 IENTER I 


Next, the user is asked to enter the first recurrence interval. 

Pre.. Fl aDd theD press ENTER to ezit 

HOW MANY PAIRS OF DISCHARGES AND 
RECURRENCE INTERVALS DO YOU WISH TO ENTER? 8 

ENTER RECURRENCE INTERVAL NUMBER 1 2IENTER I 

5-16 



The user is then asked to enter the corresponding discharge. 

Preis Fl and then pre.. ENTER to uit 

HOW MANY PAIRS OF DISCHARGES AND 
RECURRENCE INTERVALS DO YOU WISH TO ENTER1 8 
ENTER RECURRENCE INTERVAL NUMBER 1 
ENTER 2 . YEAR DISCHARGE 10 IENTER I 

After the discharge is entered, the screen is cleared and the first pair of data is 
displayed; the user is prompted to enter the next recurrence interval. 

Pre•• F1 and then pre.. ENTER to ezit 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 

ENTER RECURRENCE INTERVAL NUMBER 2 51 ENTER I 

5-17 



Again, the user is asked to enter the corresponding discharge. 

c· 

P~&I 
F1 alld thell P~&I 
ENTER to nit 

DATA PAIR RECUllRENCE INTERVAL DISCHARGE 
2 10 

ENTER RECURRENCE INTERVAL NUMBER 2 5 

ENTER 5 . YEAR DISCHARGE 691 ENTER I 

The same procedure continues until all of the data have been entered 
(six pairs in this example). 

P~.. F1 8lld thell P~" 
ENTER to nit 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 
2 5 69 

ENTER RECURRENCE INTERVAL NUMBER 3 10 IENTER I 

5-18 



Pre.. Fl a..d the.. p..... ENTER to exit 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 
2 1\ 

89 

ENTER RECURRENCE INTERVAL NUMBER 3 10 

ENTER 10 . YEAR DISCHARGE 1911 ENTER I 

P..... Fl a..d the.. pre.. ENTER to exit 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 
2 1\ 69 
3 10 191 

ENTER RECURRENCE INTERVAL NUMBER 4 20 IENTER I 

5-19 



,. 


Press F1 alld thell prou ENTER to osit 

DATA PAIR RECURRENCE Im'ERVAL DISCHARGE 
2 10 
2 5 69 
3 10 191 

ENTER RECURRENCE Im'ERVAL NUMBER 4 20 

ENTER 20 • YEAR DISCHARGE 441 IENTER I 

Press F1 &lid thell presl ENTER to ezit 

DATA PAIR RECURRENCE lNTERVAL DISCHARGE 
2 10 
2 569 
3 10 191 
4 20 441 

ENTER RECURRENCE Im'ERVAL NUMBER 5 so IENTER I 

5-20 



P",ss Fl and the" p",ss ENTER to e:rit 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 
2 589 
3 10 

191 
4 20 

441 

ENTER RECURRENCE INTERVAL NUMBER 5 50 

ENTER 50 -Y.EAR DISCHARGE 

1132 IENTER I 

P",s. Fl a"d the" p"'ss ENTER to e:rit 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 
2 5 89 
3 10 191 
4 20 441 
5 50 1132 

ENTER RECURRENCE INTERVAL NUMBER 8 

100 IENTER I 

5-21 



,


P.... Fl aDd thall p..ss ENTER to azit 
DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 
2 5 69 
3 10 191 
4 20 441 
5 50 1132 
ENTER RECURRENCE INTERVAL NUMBER 6 100 
ENTER 100 . YEAR DISCHARGE 2121 ENTER I 

Note an error has been made in entering the 100-year flood discharge. The 
program notifies us of the error with the following screen. 

P.... F1 alld thall p.... ENTER to .zit 

SORRY, DISCHARGE VALUES CANNOT DECREASE WITH INCREASING RECURRENCE INTERVALS 

THE 100 . YEAR DISCHARGE ( 212 CFS ) IS LESS THAN 
THE 50 . YEAR DISCHARGE ( 1132 CFS ) 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
5 50 1132 
6 100 212 

WHICH PAIR OF DATA DO YOU WISH TO CHANGE 


DATA PAIR NUMBER 5 OR DATA PAIR NUMBER 6? 6 IENTER I 

5-22 



The user is then prompted to re-enter the recurrence interval first. 

Pre•• F1 and then pross ENTER to nit 

SORRY DISCHARGE VALUES CANNOT DECREASE WITH INCREASING RECUllRENCE INTERVALS 

THE 100 -YEAR DISCHARGE ( 212 CFS ) IS LESS THAN 
THE 50· YEAR DISCHARGE ( 1132 CFS ) 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 
5 50 1132 
6 100 212 

WHICH PAIR OF DATA DO YOU WISH TO CHANGE •. 
DATA PAIR NUMBER 5 OR DATA PAIR NUMBER 8? 8 

RE··ENTER RECURRENCE INTERVAL NUMBER 8 

100 IENTER I 

Now, the discharge value that was entered incorrectly is corrected. 

Pres. F1 acd then press ENTER to nit 

SORRY, DISCHARGE VALUES CANNOT DECREASE WITH INCREASING RECURRENCE INTERVALS 

THE 100 . YEAR DISCHARGE ( 212 CFS ) IS LESS THAN 
THE 50 -YEAR DISCHARGE ( 1132 CFS ) 

DATA PAIR RECURRENCE INTERVAL DISCHARGE 

5 50 1132 

6 100 212 

WHICH PAIR OF DATA DO YOU WISH TO CHANGE. 

DATA PAIR NUMBER 5 OR DATA PAIR NUMBER 8? 8 

RE··ENTER RECUllRENCE INTERVAL NUMBER 8 100 

ENTER 100 -YEAR DISCHARGE 2120 IENTER I 

5-23 



The user may review the final set of data pairs. If no revisions are necessary, 
an "N"is entered. 

P..... F1 &lid thin p..... ENTER to exit 

,


DATA PAIR RECURRENCE INTERVAL DISCHARGE 
2 10 
2 5 89 
3 10 191 
• 20 "1 
5 50 1132 
8 100 2120 

DO YOU WISH TO CHANGE ANY DATA (YIN>? N IENTER I 

Entering "N" completes the data entry requirements. When the computation 
of the flood-frequency curve is complete, the statistics are displayed and the 
cu rve is drawn. 

FLOOD FREQUENCY CURVE 

10..oo
.999 .99 .9 .5 .1 .01 .001 

5-24 



As in the first example, when the computations of the contour widths are 
completed, a message appears at the bottom of the screen. 

FLOOD FREQUENCY CURVE 

10..oo 
I!!""T""'-~""'-""''''''-''''!!I 


MEAN -.9988703 
STD DEV = 1.000388 
SKEW =.0 
COR COEF =.9999992 


10 

.999 .99 .9 .5 .1 .01 .001 

•••••••• PRESS ENTER TO CONTINUE ••..••.. 1 ENTER I 


EXAMPLE NUMBER 2 

~3.5\ 
\
\ 

SINGLE·CHANNEL REGION 

4.5 
\ 
5.5~\ 
\,

6'~ 
.-...-)\ ~ 
, 

I 
'II I 
'II I 

I

I' ' 

,I 
I' ,II 

<L 
I 

I 
I ,II 

1.5 
I 
, 

I 
I 
___DEPTH ,I

,I

----VELOCITy 

.5 

•••••••• PRESS ENTER TO CONTINUE •••••••• \ ENTER I 

By pressing JENTER l the user clears the screen and the picture is drawn 
again-thIs tIme WI ~ 
the boundaries for the zones for the multiplecflannel 
region shown. 

5-25 


EXAMPLE NUMBER 2 

MULTIPLE·CHANNEL REGION 

4.5 

I 
\ 

\ 

\ 
\ 

\ 

1 
I 
I 
I 

,I

I ,I

I 
I 

__---'DEPTH 

··············_··VELOCITY 

........ PRESS ENTER TO CONTINUE ••...... 1 ENTER I 


.5 

The run is complete. The user is asked again if he/she wants to view the 
output. The answer is "No," so an "N" is entered. 

DO YOU WISH TO VIEW SOME OUTPUT DATA (YIN>? N IENTER I 

The next prompt asks if the user wishes to print the output. 

5-26 



DO YOU WISH TO PRINT THE OUTPUT (Y/Nl? Y IENTER I 

By entering "y", the user sends the output file to the printer. 
Before exiting, the program will ask the user if another run is to be made. 


DO YOU WISH TO MAKE ANOTHER RUN (Y/Nl? NIENTER I 

By entering "Nn 
, the user returns control to DOS. 

5-27 



OUTPUT FROM EXAMPLES 

The output data forthe two examples are presented in Figures 5-3 to 5-7. 

To demonstrate that the contour widths given in the last column (Width) of the 
second and third pages of the output do indeed satisfy Equation (2.31) (or Equation 

(2.37) for the multiple-channel region), the calculation for the contour width of 623 
feet corresponding to an energy of 1.5 feet will be reproduced. Note that at critical 
flow the depth of water in a wide rectangular channel is two-thirds of the energy. 
Thus, the depth in the second column of the output is 1.0 foot, as shown in Figure 
5-4. The discharge associated with alluvial fan flooding with an energy of 1.5 feet in 
the single-channel region is determined below (see Equation 2.18). 
q= 274.4D2.5 

= 274.4(1.5)2.5 

= 756 cfs (5.1) 

This discharge is given in the third column of the output (Figure 5-4.). 

The fourth and fifth columns of the output are the probabilities that 756 cfs is 
exceeded at the apex b~ 
the discharge, Q, and by 8.3176 times the first power of the 
discharge, 8.3176 Ql.OO , respectively. Table 5-1 shows the information used by the 
program to calculate those probabilities. 

5-28 



EXAMPLE NUMBER 1 

AVULSION FACTOR = 1.0000 

FLOOD FREQUENCY CURVE DEFINED BY MEAN, STANDARD DEVIATION, AND SKEW 

MEAN = 1.000000 

STANDARD DEVIATION = 1.000000 
SKEW = 0.0 

SUMMARY OF DISCHARGES: 

10-YEAR DISCHARGE = 191 

50-YEAR DISCHARGE = 1132 
100-YEAR DISCHARGE = 2120 
500-YEAR DISCHARGE = 7554 

STATISTICS AFTER TRANSFORMATION OF Y=LOGCQ) TO Z=0.9200+LOGCQ) 

MEAN OF Z = 1.920000 
STANDARD DEVIATION = 1.000000 
SKEW = 0.000000 
TRANSFORMATION CONSTANT = 3.819044 

Figure 5-3. Page 1of Output for Example Number 1 

5-29 



z· 

EXAMPLE NUMBER 1 PAGE 2 

SINGLE-CHANNEL REGION 

PROBABILITY OF DISCHARGE 
BEING EXCEEDED AT THE 
ENERGY DEPTH DISCHARGE APEX BY: WIDTH 
(FT) (FT) (CFS) 1.0000 (FT) 
Q 8.3176 Q 
0.5 0.3 49 0.24912 0.59255 2129 
1.5 1.0 756 0.03083 0.17341 623 

PROBABILITY OF DISCHARGE 

BEING EXCEEDED AT THE 
VELOCITY DEPTH DISCHARGE APEX BY: WIDTH 
(FT/SEC) (FT) (CFS) 1.0000 (FT) 

Q 8.3176 Q 

3.5 0.4 68 0.20348 0.53549 1924 
4.5 0.6 238 0.08696 0.32512 1166 
5.5 0.9 649 0.03560 0.18853 676 
6.5 1.3 1496 0.01556 0.10608 366 
Figure 5-4. Page 2 of Output for Example Number 1 

5-30 



EXAMPLE NUMBER 2 

,


AVULSION FACTOR = 1.0000 

FLOOD FREQUENCY CURVE DEFINED BY LEAST-SQUARES FIT OF DATA 

RETURN INTERVAL INPUT DISCHARGE 
(YEARS) (CFS) 
2 10 
5 69 
10 191 
20 441 
50 1132 
100 2120 

MEAN = 
STANDARD DEVIATION = 

SKEW = 

SUMMARY OF DISCHARGES: 

10-YEAR DISCHARGE = 

50-YEAR DISCHARGE = 
100-YEAR DISCHARGE = 
500-YEAR DISCHARGE = 

STATISTICS AFTER TRANSFORMATION OF 

MEAN OF Z = 
STANDARD DEVIATION = 

SKEW = 

TRANSFORMATION CONSTANT = 

BEST FIT DISCHARGE 
(CFS) 


10 
69 
191 
441 
1131 
2119 

0.998870 

1.000388 

0.0 

191 
1131 
2119 
7553 


Y=LOG(Q) TO Z=0.9207+LOG(Q) 

1.919584 
1.000388 
0.000000 
3.816320 


Figure 5-5. Page 1 of Output for Example Number 2 

5-31 



EXAMPLE NUMBER 2 PAGE 2 

SINGLE-CHANNEL REGION 

PROBABILITY OF DISCHARGE 
BEING EXCEEDED AT THE 
ENERGY DEPTH DISCHARGE APEX BY: WIDTH 
(FT) (FT) (CFS) 1.0000 (FT) 
Q 8.3313 Q 
0.5 0.3 49 0.24885 0.59235 2127 
1.5 1.0 756 0.03080 0.17340 623 

PROBABILITY OF DISCHARGE 

BEING EXCEEDED AT THE 
VELOCITY DEPTH DISCHARGE APEX BY: WIDTH 
(FTjSEC) (FT) (CFS) 1. 0000 (FT) 

Q 8.3313 Q 

3.5 0.4 68 0.20322 0.53532 1922 
4.5 0.6 238 0.08688 0.32503 1165 
5.5 0.9 649 0.03557 0.18852 676 
6.5 1.3 1496 0.01555 0.10609 366 
Figure 5-6. Page 2 of Output for Example Number 2 

5-32 



,


EXAMPLE NUMBER 2 PAGE 3 

MULTIPLE-CHANNEL REGION 

SLOPE = 0.0850000 
N-VALUE = 0.0500000 

PROBABILITY OF DISCHARGE 
BEING EXCEEDED AT THE 
ENERGY DEPTH DISCHARGE APEX BY: WIDTH 
(FT) (FT) (CFS) 1.0000 (FT) 
Q 8.3313 Q 
0.5 0.3 426 0.05206 0.24163 3277 

PROBABILITY OF DISCHARGE 
BEING EXCEEDED AT THE 
VELOCITY DEPTH DISCHARGE APEX BY: WIDTH 
(FTjSEC) (FT) (CFS) 1.0000 (FT) 
Q 8.3313 Q 
4.5 0.4 925 0.02465 0.15351 2085 

Figure 5-7. Page 3 of Output for Example Number 2 

5-33 



Table 5-1. Probabilities and Corresponding Values of k (Skew = 0) 

Probability 
0.9999 
0.9995 
0.9990 
0.9980 
0.9950 
0.9900 
0.9800 
0.9750 
0.9600 
0.9500 
0.9000 
0.8000 
0.7000 
0.6000 
0.5704 
0.5000 
0.4296 
0.4000 
0.3000 
0.2000 
0.1000 
0.0500 
0.0400 
.0.0250 
0.0200 
0.0100 
0.0050 
0.0020 
0.0010 
0.0005 
0.0001 

k 

-3.71902 
-3.29053 
-3.09023 
-2.87816 
-2.57583 
-2.32635 
-2.05375 
-1.95996 
-1.75069 
-1.64485 
-1.28155 
-0.84162 
-0.52440 
-0.25335 
-0.17733 
0.00000 
0.17733 
0.25335 
0.52440 
0.84162 
1.28155 
1.64485 
1.75069 
1.95996 
2.05375 
2.32635 
2.57583 
2.87816 
3.09023 
3.29053 
3.71902 

Note that loglO (756) is 2.87852 and is, therefore, 1.87852 standard deviations (of 1.0) 
from the mean 1.0. Thus, the probability of 756 efs being exceeded at the apex by Q 
is 0.03083, and is computed as follows: 

1.95996 -1.87852 )

P(Q >756) = ( 9 (0.040 -0.025) + 0.025 

1.95 96-1.75069 
= 0.03083 (5.2) 
This is given in the fourth column of the output, shown in Figure 5-4. 

5-34 



Also note that lag.lQ (756) is 0.95852 standard deviations (of 1.0) from the mean 1.92 
(See Equation 2.1/). Thus, the probability of 8.3176 Q1.0000 exceeding 756 cfs at the 
apex is 0.17342, and is computed as follows: 

1.28155-0.95852 )

P(8.3176 Ql.OOOO > 756)= (0.2000-0.1000) +0.1000

( 

1.28155 -0.84162 

(5.3)
=0.17342 

This is given in the fifth column of the output. (The 0.00001 difference between the 
value above and that shown in Figure 5-4 arises from rounding the discharge value 
and its base 10 logarithm.) The latter probability could also be calculated using the 
change in scale of Q. That is, 

P(8.3176 Ql.OOOO > 756) = P(Q> 756/8.3176 = 90.89) 

(5.4) 
Because laglo (90.89) is 1.95852 and is, therefore, 0.95852 standard deviations (of 1.0) 
from the mean 1.0, Equation (5.3) holds for the rescaled discharge and the floodfrequency 
curve defined at the apex for Q. 

Equation (5.3) defines the probability denoted P(Z>lagloq·) in Equation (2.31). The 
transformation constant C in Equation (2.31) is defined by Equation (2.16) 

2

0.92}l + 0.420

C =e 

=eO.92 + 0.42 

= 3.819044 (5.5) 

This is given on the first page of the output (see Figure 5-3). The avulsion factor in 
the examples is 1.0. 

The correction for high flow values is accomplished by calculating the value of the 
discharge (denoted q ,in equation (2.31)) that would create a channel as wide as the 
contour width, 623 feet. By using Equation (2.6), qw is determined by 

(
W )2.5 

qw = 9.408 

5-35 


=( 623 )'2.5 

9.408 

(5.6)
= 35684 d's 
The k-values that correspond to qw ' and the rescaled qware 

(5.7)
loglO(35684) -1.0 = 3.55247 
and 

loglO(~5684) 
-1.92 = log10 (35684/8.3176) -1.0 

= 2.63247 (5.8) 
Therefore, the two probabilities, P(Q>q",J and P(Z> loglO q,) in Equation (2.31) are 
3.71902-3.55247 )

P(Q>q ) = ( . (O.0005-0.000I) +0.0001 

w 3.71902 -3.29053 
=0.00026 (5.9) 
and 

2.87816-2.63247 )

= ( (0.0050-0.0020) +0,0020

2.87816-2.57583 

= 0.00444 (5.10) 

Thus, the probability that a point on the contour at which the area subject to 
flooding is 623 feet wide will be inundated by a flood discharge of 756 cfs or more is 
from Equation (2.31) 

5-36 



=(9.408)(1.0)(3.819044) [0.17342-0.00444] + 0.00026 
623 

= 0.01001 (5.11 ) 

Carrying out the calculations to a precision of eight significant figures yields a value 
of 0.01000021 for the probability. A calculation without the large flow correction, 
but with the same precision, yields a probability of 0.01000068. 

DELINEATING FLOOD INSURANCE ZONE BOUNDARIES 

FEMA designates areas subfect to 100-year alluvial fan flooding as Zone AD, with 
100-year flood depths and velocities shown. The depths shown are summations of 
the pressure heads and velocity heads. Depths are rounded to the nearest whole 
foot and velocities are rounded to the nearest foot per second (fps). Thus, the area 
subjeetto alluvial fan flooding with 100-year flood depths between 1.5 and 2.5 feet 
and 100-year flood velocities between 5.5 and 6.5 fps is labeled: 

ZONEAO 

(DEPTH 2) 
eVE LDCITY 6 FPS) 


The net output data of the FAN program are the energies (labeled DEPTH on the 
FIRM), velocities, and the corresponding widths. The latter are the widths of the 
area subject to alluvial fan flooding at which the corresponding energy or velocity 
has a 1-percent chance of being exceeded in any given year. To delineate the flood 
insurance zone boundaries, the elevation at which the area subject to alluvial fan 
flooding has a width equal to the width in the output must be determined. To 
shorten the terminology, that width is referred to as the contour width (because it is 
measured along a contour). 

The contour width is not the length of the contour, including every bend and wind, 
between the boundaries of the area subject to flooding. Instead, it is the length of a 
"smoothed" contour that has the same general alignment as the "true" contour. 
Figure 5-8 shows the alluvial fan used in the previous examples with the boundaries 
of the area subject to flooding. Figure 5-9 shows the same fan with "smoothed" 
contours for width measurements. 

The output data from Example Number 1will be used to demonstrate. To delineate 
the boundary between the zones where the 100-year flood depths are 1 foot and 
less than 1 foot, the elevation at which the contour width is 2,129 feet is located. To 
delineate the boundary between the zones where the 100-year flood depths are 1 
foot and 2 feet, the elevation at which the contour width is 623 feet is located. The 
zone boundaries are delineated at those elevations using the alignments suggested 
by the smooth contours on Figure 5-9. The area between those boundaries will be 
labeled "DEPTH 1." The area above the 623-foot-long boundary will be labeled 
"DEPTH 2." Note that, in the example, the 100-year flood discharge at the apex is 
2,120 cfs. Therefore, the maximum 100-year flood depth is 2.26 feet and, so, the 
maximum depth labeled is 2 feet. (See Condition 2.23.) 

5-37 


/I 

/{ 

/I 

/\ 

\ 

1" =1000' 

"


"


"


"'


" 

Figure 5-8. Alluvial Fan Boundaries 

5-38 


/I 
/I 
II I 

c-/~ 
I~ 
I~ 
\ 


<OO-, 


II 

/ ~ 


/~/ 
./~ 
/~ 
/~ 
\ 1" =1000' 

'OOO~

\ 

"


"


"


"


/ "-,

/

/

/

/

/
/ 


Figure 5-9. "Smooth" Contours for Width Measurements 

5-39 


The delineations of the depth zone boundaries for this single-channel region are 
shown on Figure 5-10. The same procedure is followed to delineate the velocity 
zone boundaries. The velocity zone boundaries subdivide the depth zones. The end 
result is a map depicting a series of regions defined by pairs of 100-year flood depths 
and velocities. Figure 5-11 shows that result with the appropriate labels in the flood 
insurance zones. 

In multiple-channel regions, the widths resulting from the multiple-channel analysis 
are used. In Example Number 2, the multiple-channel region is the area with 
elevations below 1,180 feet. Because the area subject to flooding is 1,050 feet wide 
at that elevation, all widths greater than 1,050 feet are taken from the multiplechannel 
output data. For example, the boundary between the 4-fps and 5-fps zones 
is where the area subject to flooding is 2,085 feet wide--not 1,165 feet wide (singlechannel 
region); the boundary between the 1-foot depth zone and areas where 100year 
flood depths are less than 0.5 foot is where the area subject to flooding is 3,277 
feet wide--not 2,127 feet wide (single-channel region). Figure 5-12 shows the results 
corresponding to Example Number 2 with bifurcation points at elevation 1,180 feet. 

5-40 



I/ 
II I 

/ .)--DEPTH = 2 FT 

,-/~ 


/I 

I\ 

/\ 

I\ 

/I 

/I 
I J 
I/ 

/ / 

/ DEPTH =1 FT/ 

I I 
II \ 
I\ 

1" =1000' 

/ \ 

I\ 

/ 

\

/ 

\.

I 

\

I 

\

I 

\

/\ 

I \. 

I \. 

\.

I 

"


I 

"


I 

" "


II ""


"


I 

"


I "I 


/ " ", 

I 
I 


/

I 

/ 

Figure 5-10. Depth Zone Boundaries (Single-Channel Region) 

5-41 


I
/


/"1 ZONE AO 
ZONE AO ~ 
(DEPTH 2 FT) 
~ 
(DEPTH 2 FT) ----t--(VELOCITY 7 FPS) 
(VELOCITY 6 FPS) ~ 
\ 

/~ZONEAO

\ (DEPTH 1 FT)

I

I

(VELOCITY 6 FPS) 

\ 


I ZONE AO \ 

(DEPTH 1 FT) I 
(VELOCITY 5 FPS) 1 

.

I

/ 

I


J 


/~/ 


/ 


/ 

ZONEAO I 

I 

(DEPTH 1 FT) \ 
(VELOCITY 4 FPS) 

\ 

/


I

I

I 

/ 


1" = 1000'


\

I

/ 

--:.---ZONEAO 

(DEPTH 1 FT) 

I 


/ 

\ 

(VELOCITY 3 FPS) 

/ 

'\

/ 

\ 

\ 

ZONE B 

\ 

I 

I 

/ 


\ 

'\ 

'\ 

'\ 

"""


""


IIII 
f 

I 


"" """"-, 


I 


IJ 


/

/

I 


/

/

/

/ 

Figure 5-11. Flood Insurance Zones (Single-Channel Region) 

5-42 


/ \ 
\ 
\/
/
/ 
/ \ 
\ 
\/
/
/ 
/I 

/ '"1 ZONE AO 
ZONE AO ~ 
(DEPTH 2 FT) 

(DEPTH 2 FT) ~ 
(VELOCITY 7 FPS) 
(VELOCITY 6 FPS) ~ 
\ 

/~ZONEAO 


/ \ (DEPTH 1 FT)I \ (VELOCITY 6 FPS) 

I\ 

I I ELEVATION OF 
/ I--BIFURCATION 
I J POINT 

// (1180 FT.) 

/I 

// 

II ZONE AO /
I (DEPTH 1 FT) I
I (VELOCITY 5 FPS) \ 


1" =1000' 

\

/ 

'\

I ZONE AO 

\ 

(VELOCITY 4 FPS) 
(DEPTH 1 FT) 

\

/ 
/ 



\

/ 

'\ 

'\

I 

/'\ 
"


I 

"


I ZONE B "


I "


I '""


I "


"


II "-'"'",

I 

I 

/

/

/

/ 

Figure 5-12. Flood Insurance Zones (Multiple-Channels Region) 

5-43 


SECTION 6 -REFERENCES 


1. 
David R. Dawdy, "Flood Frequency Estimates on Alluvial Fans," in Journal of 
Hydraulics Division, ASCE, Proceedings, Vol. 105, No. HYII, 1979, pp 1407-1413. 
2. 
DMA Consulting Engineers, Alluvial Fan Flooding Methodology, An Analysis, 
prepared for Federal Emergency Management Agency, 1985. 
3. 
Interagency Advisory Committee on Water Data, Hydrology Subcommittee, 
Bulletin No. 17B, Guidelinesfor Determining Flood Flow Frequency, 1982. 
6-1 



APPENDIX A -SUPPLEMENT TO DERIVATION 


HYDRAULIC CHARACTERISTICS: SINGLE-CHANNEL REGION 

Recall that, at critical depth, the water depth, d, in a rectangular channel is 
twice the velocity head, v2/2g. Thus, for a constant discharge, q, flowing at 
critical velocity, v, in a rectangular channel of cross-sectional area, A, and 
width, w, 

y2 ) 1 2 
d=2( 2g =gy 

(A.1 ) 
where g is the acceleration of gravity. 

Therefore, 

2 

2 q1 

w =-(A.2) 
g d3 

or, writing the width as a function of depth, 

q_ d-312

wed) = 

(A.3)
vg 

The condition 

dw 

-= -200 

(A.4)
dd 

A-1 



where d denotes differentiation with respect to depth, yields

dd 
dw = _~ 
(~)d-512 


(A.S)
dd vg .2 
That is, Condition (A.4) is satisfied when 


= 0 .07056 q2l5 (A.G) 
Writing the width as a function of discharge [Equation (A.3)] yields 
q_ d-312

w(q) = 

Vg 
= q_[3q ] -3/5 
V g 400V g 
= 9.408 q2l5 

(A.7) 
The discharge associated with an energy depth of D feet is the discharge 
corresponding to a water depth of d=2D/3 feet in Equation (A.G): 

320 [( q )'(3) 1]215 (A.S)

= vg2 200 
That is, 


-(3 )-3.5 

5

q= 200V g -02. 

2 

5 (A.9)

= 274.4 0 2. 

A-2 



Similarly, the discharge associated with a velocity, v, is the discharge 
corresponding to a water depth of d =2(v2/2g) feet in Equation (A.G): 

,


(A.10) 
That is, 

400)· 5 

q= ( 
-y 
3 g2 

= 0.1289y5 
(A.11 ) 

A.2 HYDRAULIC CHARACTERISTCS: MULTIPLE-CHANNEL REGION 
Assume that the depth and velocity of floodflows can be estimated with 
Manning's equation with the frktion slope set equal to the slope of the alluvial 
fan (Le., a normal depth approximation). Thus, we define the relationship 
between depth and discharge as 

1.486 213 112 
q= --AR s 
(A.12)

n 

where s is the slope of the alluvial fan, R is the hydraulic radius of the channel, 
and A is the cross-sectional area of that part of the channel under water. 
Assuming a wide rectangular channel, we can approximate the hydraulic 
radius, R, by the depth of water, d. Writing the width as 

w(q)=(3.8)(9.408Iq2'5 =35.7504 q2'5 
(A.13) 

and the area as width times depth yields 

1.486 
q =-w(q) d513 s1l2 
n 

1.486 
= --35.7504q2'5d5l3 s1J2 
n 

53.1251 215 513 112

= qds 
(A.14) 

n 

A-3 


Solving for d yields 

d[ n 315 -112 ]315 
= 53.1251 q 5 

= 0.0922 nO.6 5-0.3 qO.36 (A.1S) 

Manning's equation for velocity, v, is 

1.486 213 1/2
v= --R 5 (A.16) 

n 

Making the same substitution for R as before yields the depth-velocity 
relationship 

1.486 213 112
v=--d 5 (A.17) 

n 
Using Equation (A.1 5) for d yields 

(A.18) 
The velocity head can be computed from Equation (A.18): 

(A.19) 
A-4 



,Therefore, 
the energy depth, D, in the multiple-channel region is 
v2 
D=d+ 2g 
A.3 DERIVATION OF EQUATION (2.11) 
Rearranging 
0.01 PCH= = 1) = f 
IX> 
y 100 
eO.92y Ak(y_m)k-l 
-A(v-m) d9.408 we· v f(k) . 
(A.21 ) 
to 
9.408 
0.01 =-W 
suggests writing 
( )k-l y-m 
nk) 
(A.22) 
as 
which is the same as 
(A (
A k 
0.92) .\k e -(A-O.92)iy-ml eO.92m 
0.92)k 
eO.92m hk 
___ 
(A-0.92)k 
Defining the constant 
(h_0.92)k e-<.\-O.92Hy-ml 
c=--(.\ 
0.92)
k 
(A.23) 
A-S 


and the new parameter 
A = A-0.92 (A:24) 
yields 
0.01 
9.408C fQD =-W 
Y100 
(A,)k(y_m)k-l 
fCk) 
-Aly-ml d 
e y 
(A.25) 
A.4 DERIVATION OF EQUATION (2.15) 
Start with 
( 
Y-I1)2-t -
f 
QD O.92y C1 
0.01= 9.408 _e_ _e_-=~dY 
YlOO W ov'2n 
Expand the square, rearrange, and add and subtract 
[ 
(0.920)2 ] 
0.9211+ --2to 
the exponent. 
That is, the exponent can be written 
(A.26) 
= 
1 2 
--2[y -(ll+ 0.9202)] + 0.9211 + 0.4202 
20 
(A.27) 

A-6 



Defining the constant 

2 
C 0.92~ 
+ 0.420 
=e (A.28) 
and the new mean 
lI' = 1I + 0.9202 (A.29) 
yields 
2 
0.01 
9.408 C fr 
-~Y~~') 
e-------dy 
(A.3D) 
W YlOO ov'2n 

A-7 



APPENDIX B -LISTING 

Table of Contents for Program FAN 

Page 

PEARSN.BAS B-3 

Display Introductory Screen B-3 
Open File to Store Input B-3 
Assignk-Values forNon-NegativeSkews.............................. B-4 
Assign k-Values for Negative Skews B-12 
Terminate Run If F1 Has Been Entered B-12 

FANINP.BAS B-13 

Enter Name of Alluvial Fan B-13 
Input Multiple-Channel Data B-14 
Input Avulsion Factor B-15 
Choose Option for Defining Flood-Frequency Curve B-15 
Option 1-Enter Statistics B-16 
Option 2 -Enter Pairs of Data B-17 
Find Skew That Gives Best Fit B-18 
Find Probability and k-Value for Given Recurrence Interval 
and Discharge B-18 
Find Least-Squares Fit of Data B-19 
Check 100-Year Discharge Value B-19 
Write Input to Input File B-20 
Check Recurrence Interval B-21 
Check That Discharge is Greater Than Zero B-21 
Order Data by Recurrence Interval and Check that 
Discharges Do Not Decrease B-22 
Terminate Run If F1 Has Been Entered B-23 

FANRUN.BAS B-24 

Check for Graphics Capability B-24 
Read Input B-24 
Main Program B-25 
Transform Random Variable B-26 
ComputeContourWidthsforDepthZones............................ B-26 
Compute Discharge Corresponding to Depth in 
Multiple-Channel Region B-27 
Compute Contour Widths for Velocity Zones B-28 
Compute Probability B-29 
Adjust Contour Width B-29 
Write Flood-Frequency Output B-30 
Write Transformation Output B-31 
Write Depth-and Velocity-Zone Output B-32 
Draw Flood-Frequency Curve B-34 
Draw Fan B-35 
Option to View Output B-36 
Display Flood-Frequency Output B-37 

B-1 



Table of Contents for Program FAN (Cont'd) 

,


Page 

Display Transformation Output . B-38 
Display Depth-Zone Output (Single Channel) . B-39 
Display Velocity-Zone Output (Single Channel) . B-40 
Display Depth Zone Output (Multiple Channels) . B-41 
Display Velocity Zone Output (Multiple Channels) . B-42 
Optionto Print Output . B-42 

AGAIN.BAS B-43 

B-2 



1 REM 

2 REM PPPPPPPP EEEEEEE AAA RRRRRRRR SSSSSSSS NNN NNN 

3 REM PPP PPP EEE AAAAA RRR RRR SSS NNNN NNN 

4 REM PPP PPP EEE AAA AAA RRR RRR SSS NNNNNN NNN 

5 REM PPPPPPPP EEEEEE AAA AAA RRRRRRRR SSSSSSSS NNN NNN NNN 

6 REM PPP EEE AAAAAAAAAAA RRR RRR SSS NNN NNNNNN 

7 REM PPP EEE AAA AAA RRR RRR SSS NNN NNNN 

8 REM PPP EEEEEEE AAA AAA RRR RRR SSSSSSSS NNN NNN 

9 REM 

10 REM Portions (c) Copyright

11 REM Microsoft Corporation

12 REM 1982-1988 

13 REM All Rights Reserved 

14 REM 

15 REM 

16 REM ALLUVIAL FAN /-Michael Baker Jr., Inc. -JULY 1990 

17 REM 

18 REM 

19 COMMON SHARED PC), K(), KG(), KO()

20 DIM P(30), K(83, 30), KG(30), KO(30) 

21 REM 

22 ON KEY(l) GOSUB 10000 

23 KEY(l) ON 

24 CLS : KEY OFF: SCREEN 0: COLOR 10 

25 LOCATE 6, 20 

26 PRINT "ALLUVIAL FAN FLOODING COMPUTER PROGRAM" 

28 LOCATE 9, 20· 
30 PRINT "In response to prompts, enter data one " 
31 LOCATE 10, 20 


32 PRINT "value at a time, then press enter." 
34 LOCATE 14, 20 


35 PRINT "Answer yes-or-no questions with the" 
36 LOCATE 15, 20 
37 PRINT "corresponding letters (i.e., Y or N).n 
38 LOCATE 20, 20 


40 REM 
52 REM 
54 REM ************************************************************ 

56 REM OPEN OUTPUT FILE 
58 REM *****************************************************~****** 


60 REM 
62 OPEN "FAN. IN" FOR OUTPUT AS '1 
63 LOCATE 20, 20 
64 COLOR 18 
66 PRINT "PEARSON TYPE-III TABLES BEING LOADED" 

B-3 



9000 REM 
9002 REM *********************************************************.-~ 


9004 REM ASSIGN K VALUES FOR PEARSON TYPE 3 
9006 REM ************************************************************ 
9008 REM 
9012 P(O) = .9999: P(l) = .9995: P(2) = .999: P(3) = .998: P(4) = .995: 


P(5) = .99: P(6) = .98 

9014 P(7) = .975: P(8) = .96: P(9) = .95: P(10) = .9: P(ll) = .8: P(12) 
= .7: P(13) = .6 

9016 P(14) = .5704: P(15) = .5: P(16) = .4296: P(17) = .4: P(18) = .3: P 

(19) = .2: P(20) = .1 
9018 P(21) = .05: P(22) = .04: P(23) = .025: P(24) -.02: P(25) = .01: P 

(26) = .005: P(27) = .002 
9020 P(28) = .001: P(29) = .0005: P(30) = .0001 

9022 K(O, 0) = -3.71902: K(l, 0) = -3.50703: K(2, 0) = -3.29921: K(3, 0) = -3.09631: K(4, 0) = ~2.89907: 
K(5, 0) --2.70836: K(6, 0) = -2.52507 
9024 K(O, 1) = -3.29053: K(l, 1) = -3.12767: K(2, 1) = -2.96698: K(3, 1) 

= -2.80889: K(4, 1) = -2.6539: K(5, 1) = -2.50257: K(6, 1) = -2.35549 
9026 K(O, 2) = -3.09023: K(l, 2) --2.94834: K(2, 2) --2.80786: K(3, 2) = -2.66915: K(4, 2) = -2.53261: K(5, 2) --2.39867: K(6, 2) = -2.2678 
9028 K(O, 3) = -2.87816: K(l, 3) = -2.75706: K(2, 3) = -2.63672: K(3, 3) 

= -2.51741: K(4, 3) = -2.39942: K(5, 3) = -2.28311: K(6, 3) = -2.16884 

9030 K(O, 4) = -2.57583: K(l, 4) = -2.48187: K(2, 4) = -2.38795: K(3, 4) 

= -2.29423: K(4, 4) = -2.20092: K(5, 4) = -2.10825: K(6, 4) = -2.01644 

9032 K(O, 5) = -2.32635: K(l, 5) = -2.25258: K(2, 5) = -2.1784: K(3, 5) 
= -2.10394: K(4, 5) --2.02933: K(5, 5) --1.95472: K(6, 5) = -1.88029 
9034 K(O, 6) = -2.05375: K(l, 6) = -1.99973: K(2, 6) = -1.94499: K(3, ~l 


= -1.88959: K(4, 6) = -1.83361: K(5, 6) = -1.77716: K(6,6) = -1.720: 

9036 K(O, 7) = -1.95996: K(l, 7) = -1.91219: K(2, r) = -1.8636: K(3, ~J 


= -1.81427: K(4, 7) = -1.76427: K(5, 7) = -1.71366: K(6, 7) = -1.66253 

9038 K(O, 8) = -1.75069: K(l, 8) = -1.7158: K(2, 8) = -1.67999: K(3, 8) = -1.64329: K(4, 8) = -1.60574: K(5, 8) = -1.5674: K(6, 8) --1.5283 
9040 K(O, 9) = -1.64485: K(l, 9) = -1.61594: K(2, 9) = -1.58607: K(3, 9) 

= -1.55527: K(4, 9) = -1.52357: K(5, 9) = -1.49101: K(6, 9) = -1.45762 

9042 K(O, 10) = -1.28155: K(l, 10) = -1.27037: K(2, 10) = -1.25824: K(3, 

10) = -1.24516: K(4, 10) = -1.23114: K(5, 10) = -1.21618: K(6, 10) = -1. 

20028 
9044 K(O, 11) = -.84162: Ke1, 11) = -.84611: K(2, 11) = -.84986: K(3, 11

) = -.85285: K(4, 11) = -.85508: K(5, 11) = -.85653: K(6, 11) --.85718 

9046 K(O, 12) = -.5244: K(l, 12) = -.53624: K(2, 12) = -.54757: K(3, 12) 

= -.55839: K(4, 12) = -.56867: K(5, 12) = -.5784: K(6, 12) = -.58757 

9048 K(O, 13) = -.25335: K(l, 13) = -.26882: K(2, 13) = -.28403: K(3, 13) = -.29897: K(4, 13) --.31362: K(5, 13) = -.32796: K(6, 13) --.34198 
9050 K(O, 14) = -.17733: K(l, 14) = -.19339: K(2, 14) = -.20925: K(3, 14

) = -.22492: K(4, 14) = -.24037: K(5, 14) = -.25558: K(6, 14) = -.27047 

9052 K(O, 15) = 01: K(l, 15) = -.01662: K(2, 15) = -.03325: K(3, 15) = .
04993: K(4, 15) = -.06651": K(5, 15) --.08302: K(6, 15) = -.09945 
9054 K(O, 16) = .17733: K(l, 16) = .16111: K(2, 16) = .14472: K(3, 16) = 

.1282: K(4, 16) = .11154: K(5, 16) = .09478: K(6, 16) = .07791 
9056 K(O, 17) = .25335: K(l, 17) = .23763: K(2, 17) = .22168: K(3, 17) = 

.20552: K(4, 17) = .18916: K(5, 17) = .17261: K(6, 17) = .15589 
9058 K(O, 18) = .5244: K(l, 18) = .51207: K(2, 18) = .49927: K(3, 18) = 
.486: K(4, 18) = .47228: K(5, 18) = .45812: K(6, 18) = .44352 
9060 K(O, 19) = .84162: K(l, 19) = .83639: K(2, 19) = .83044: K(3, 191 = 

.82377: K(4, 19) = .81638: K(5, 19) = .80829: K(6, 19) = .7995 

9062 K(O, 20) = 1.28155: K(l, 20) = 1.29178: K(2, 20) = 1.30105: K(3, 20

) = 1.30936: K(4, 20} = 1.31671: K(5, 20) = 1.32309: K(6, 20) = 1.3285 
8-4 


9064 K(O, 21) = 1.64485: K(l, 21) = 1.67279: K(2, 21) = 1.69971: K(3, 21

) = 1.72562: K(4, 21) = 1.75048: K(5, 21) = 1.77428: K(6, 21) = 1.79701 

9066 K(O, 22) = 1.75069: K(l, 22) = 1.78462: K(2, 22) = 1.81756: K(3, 22

. ) = 1.84949: K(4, 22) = 1.88039: K(5, 22) = 1.91022: K(6, 22) = 1.93896 

9068 K(O, 23) = 1.95996: K(l, 23) = 2.00688: K(2, 23) = 2.0529: K(3, 23) 
= 2.09795: K(4, 23) = 2.14202: K(5, 23) :I: 2.18505: K(6, 23) -2.22702 
9070 K(O, 24) = 2.05375: K(l, 24) = 2.10697: K(2, 24) = 2.15935: K(3, 24

) = 2.21081: K(4, 24) = 2.26133: K(5, 24) = 2.31084: K(6, 24) = 2.35931 

9072 KCO, 25) = 2.32635: KC1, 25) = 2.39961: K(2, 25) 2.47226: K(3, 25

:I: 

)= 2.54421: K(4, 25) = 2.61539: K(5, 25) = 2.68572: K(6, 25) -2.75514 
9074 KCO, 26) = 2.57583: K(l, 26) = 2.66965: K{2, 26) = 2.76321: K(3, 26) = 2.85636: K(4, 26) 2.949: K{5, 26) -3.04102: K{6, 26) = 3.13232

:I: 

9076 K{O, 27) = 2.87816: K(l, 27) = 2.99978: KC2, 27) -3.12169: K(3, 27

) = 3.24371: K(4, 27) = 3.36566: K{5, 27) = 3.48737: K{6, 27) = 3.60872 

9078 K(O, 28) = 3.09023: K{l, 28) = 3.23322: K{2, 28) -3.37703: K(3, 28

) = 3.52139: K{4, 28) = 3.66608: K(5, 28) = 3.8109: K{6, 28) = 3.95567 

9080 X(O, 29) = 3.29053: K(l, 29) = 3.45513: K(2, 29) -3.62113: K(3, 29) = 3.7882: K(4, 29) = ~.95605: 
K(5, 29) = 4.12443: K(6, 29) -4.29311 
9082 K{O, 30) = 3.71902: K(l, 30) 3.93453: K(2, 30):1: 4.15301: K(3, 30

:I:

) = 4.37394: K{4, 30) = 4.59687: K{5, 30) = 4.82141: K{6, 30) -5.04718 

9084 X{7, 0) = -2.35015: K{8, 0) = -2.18448: K{9, 0) = -2.02891: K{10, 0

) = -1.8841: K{ll, 0) = -1.75053: K{12, 0) = -1.62838: K(13, 0) = -1.5175 

2 
9086 X(7, 1) = -2.21328: K{8, 1) -2.07661: K{9, 1) --1.94611: K(10, 1) = -1.82241: K{ll, 1) = -1.70603: 
:I: 
K{12, 1) = -1.59738: K{13, 1) --1.496 
73 
9088 XC7, 2) = -2.14053: K{8, 2) --2.01739: K(9, 2) --1.89894: K(10, 2 
)= -1.78572: K{ll, 2) = -1.67825: K{12, 2) = -1.57695: KC13, 2) • -1.482 
16 
9090 X(7, 3) = -2.05701: K{8, 3) = -1.94806: K(9, 3) = -1.84244: K(10, 3 
)= -1.74062: K(ll, 3) = -1.64305: K(12, 3) = -1.55016: K(13, 3) • -1.462 
32 
9092 K(7, 4) = -1.9258: K(8, 4) = -1.8366: K(9, 4) = -1.74919: K(10, 4) = -1.6639: K(ll, 4) = -1.5811: K(12, 4) = -1.50114: K(13, 4) --1.42439 
9094 K{7, 5) = -1.80621: K(8, 5) = -1.73271: K(9, 5) = -1.66001: K(10, 5

) = -1.58838: K(11, 5) = -1.51808: K{12, 5) = -1.44942: KC13, 5) = -1.382 

67 
9096 K(7, 6) = -1.66325: K(8, 6) --1.60604: K{9, 6) = -1.54886: K(10, 6) = -1.49188: K{ll, 6) = -1.43529: K(12, 6) = -1.37929: K(13, 6) = -1.324 
12 

9098 X{7, 7) = -1.61099: K{8, 7) = -1.55914: K{9, 7) = -1.50712: K(10, 7 

)= -1.45507: K{ll, 7) = -1.40314: K{12, 7) = -1.35153: K{13, 7) = -1.300 
42 
9100 K{7, 8) = -1.48852: K{8, 8) = -1.44813: K{9, 8) = -1.4072: K(10, 8) 

= -1.36584: K{ll, 8) = -1.32414: K{12, 8) = -1.28225: K(13, 8) --1.2402 

8 
9102 K{7, 9) --1.42345: K(8, 9) = -1.38855: K(9, 9) = -1.35299: K(10, 9) = -1.31684: K{ll, 9) = -1.28019: K(12, 9) = -1.24313: KC13, 9) = -1.205 

78 . 
9104 X(7, 10) = -1.18347: K(8, 10) = -1.16574: K(9, 10) --1.14712: K(10 
, 10) = -1.12762: K(ll, 10) = -1.10726: K(12, 10) = -1.08608: K(13, 10) = 
-1. 06413 
9106 X{7, 11) = -.85703: K{8, 11) = -.85607: K(9, 11) --.85426: K{10, 1 
1) = -.85161: K{ll, 11) = -.84809: K(12, 11) = -.84369: X{13, 11) = -.838 
41 
9108 X{7, 12) = -.59615: K{8, 12) = -.60412: K{9, 12) = -.61146: X{10, 1 
2) = -.61815: K(11, 12) = -.62415: K{12, 12) = -.62944: K{13, 12) = -.634 
9110 K(7, 13) = -.35565: K{8, 13) = -.36889: K(9, 13) :I: -.38186: K {10, 1 
3) = -.39434: K(11, 13) = -.40638: K(12, 13) = -.41794: K(13, 13) = -.428 

99 

8-5 


9112 K(7, 14) = -.28516: K(8, 14) = -.29961: K(9, 14) = -.31368: K(10, 1 

4) = -.3274: K(11, 14) = -.34075: K(12, 14) = -.3537: K(13, 14) = -.3662 

9114 K(7, 15) = -.11578: K(8, 15) = -.13199: K(9, 15) = -.14807: K( 10, 1 

5) = -.16397: K(ll, 15) = -.17968: K(12, 15) = -.19517: K(13, 15) = -.210 

4 
9116 K(7, 16) = .06097: K(8, 16) -.04397: K(9, 16) -.02693: K(10, 1~ 
= .00987: K(ll, 16) = -.00719: K(12, 16) = -.02421: K(13, 16) = -.04116 
9118 K(7, 17) = .13901: K(8, 17) -.12199: K(9, 17) = .10486: K(10, 17) = 8.763001E-02: K(ll, 17) = .07032: K(12, 17) = .05297: K(13, 17) = .0356 

9120 K(7, 18) = .42851: K(8, 18) = .41309: K(9, 18) = .39729: K(10, 18) 

= .38111: K(11, 18) = .36458: K(12, 18) =.34772: K(13, 18) -.33054 
9122 K(7, 19) = .79002: K(8, 19) = .77986: K(9, 19) -.76902: K(10, 19) = .75752: K(ll, 19) = .74537: K(12, 19) = .73257: K(13, 19) = .71915 
9124 K(7, 20) = 1.33294: K(8, 20) = 1.3364: K(9, 20) = 1.33889: K(10, 20) = 1.34039: K(ll, 20) = 1.34092: K(12, 20) = 1.34047: K(13, 20) = 1.3390 

4 
9126 K(7, 21) = 1.81864: K(8, 21) = 1.83916: K(9, 21) = 1.85856: K(10, 2 
1) = 1.87683: K(ll, 21) = 1.89395: K(12, 21) = 1.90992: K(13, 21) = 1.924 
72 

9128 K(7, 22) = 1.9666: K(8, 22) = 1.99311: K(9, 22) = 2.01848: K(10, 22

) = 2.04269: K(ll, 22) = 2.06573: K(12, 22) = 2.08758: K(13, 22) = 2.1082 
3 

9130 K(7, 23) = 2.2679: K(8, 23) = 2.30764: K(9, 23) = 2.34623: K(10, 23

) = 2.38364: K(ll, 23) = 2.41984: K(12, 23) = 2.45482: K(13, 23) • 2.4885 
5 
9132 K(7, 24) = 2.4067: K(8, 24) = 2.45298: K(9, 24) = 2.49811: K(10, 24) = 2.54206: K(ll, 24) = 2.5848: K(12, 24) = 2.62631: K(13, 24) = 2.66657 
9134 K(7, 25) = 2.82359: K(8, 25) = 2.89101: K(9, 25) -2.95735: K(10, 2 

5) = 3.02256: K(11, 25) = 3.0866: K(12, 25) = 3.14944: K(13, 25) = 3.2110 
3 
9136 K(7, 26) = 3.22281: K(8, 26) -3.31243: K(9, ~6) 
= 3.40109: K(lC 
6) = 3.48874: K(11, 26) = 3.5753: K(12, 26) = 3.66073: K(13, 26) = 3.7449 
7 
9138 K(7, 27) = 3.72957: K(8, 27) = 3.84981: K(9, 27) = 3.96932: K(10, 2 

7) = 4.08802: K(11, 27) = 4.20582: K(12, 27) = 4.32263: K(13, 27) = 4.438 

39 

9140 K(7, 28) = 4.10022: K(8, 28) = 4.24439: K(9, 28) = 4.38807: K(10, 2 

8) = 4.53112: K(11, 28) = 4.67344: K(12, 28) = 4.81492: K(13, 28) = 4.955 
49 
9142 K(7, 29) = 4.46189: K(8, 29) = 4.63057: K(9, 29) = 4.79899: K(10, 2 

9) = 4.96701: K(11, 29) = 5.13449: K(12, 29) = 5.3013: K(13, 29) = 5.4673 
5 
9144 K(7, 30) = 5.27389: K(8, 30) = 5.50124: K(9, 30) = 5.72899: K(10, 3 
0) = 5.95691: K(11, 30) = 6.1848: K(12, 30) = 6.41249: K(13, 30) = 6.6398 
9146 K(14, 0) = -1.41753: K(15, 0) = -1.32774: K(16, 0) = -1.24728: K(17 
, 0) = -1.1752: K(18, 0) = -1.11054: K(19, 0) = -1.05239: K(20, 0) = -.99 
99 
9148 K(14, 1) = -1.40413: K(15, 1) --1.31944: K(16, 1) = -1.24235: K(17 
, 1) = -1.1724: K(18, 1) = -1.10901: K(19, 1) = -1.05159: K(20, 1) = -.99 
95 
9150 K(14, 2) = -1.39408: K(15, 2) = -1.31275: K(16, 2) = -1.23805: K(17 
, 2) = -1.16974: K(18, 2) = -1.10743: K(19, 2) = -1.05068: K(20, 2) = -.9 
99 
9152 K(14, 3) = -1.37981: K(15, 3) = -1.30279: K(16, 3) = -1.23132: K(17 
, 3) = -1.16534: K(18, 3) = -1.10465: K(19, 3) = -1.04898: K(20, 3) = -.9 
98 
9154 K(14, 4) = -1.35114: K(15, 4) = -1.28167: K(16, 4) = -1.21618: K,.7 
, 4) = -1.15477: K(18, 4) = -1.09749: K(19, 4) = -1.04427: K(20, 4) = -.9 
9499 
9156 K(14, 5) = ~1.31815: 
K(15, 5) = -1.25611: K(16, 5) = -1.1968: K(17, 

8-6 


5) = -1.14042: K(18, 5) = -1.08711: K(19, 5) = -1.03695: K(20, 5) = -.98 
995 
9158 K(14, 6) = -1.26999: K(15, 6) = -1.21716: K(16, 6) = -1.16584: K(17 

, 6) = -1.11628: K(18, 6) = -1.06864: K(19, 6) = -1.02311: K(20, 6) = -.9 
798 
9160 K(14, 7) --1.25004: K(15, 7) --1.20059: K(16, 7) --1.15229: K(17 

, 7) = -1.10537: K(18, 7) = -1.06001: K(19, 7) --1.0164: K(20, 7) = -.97 
468 
9162 K(14, 8) --1.19842: K(15, 8) --1.15682: K(16, 8) --1.11566: K(17 
, 8) = -1.07513: K(18, 8) = -1.03543: K(19, 8) --.99672: K(20, 8) --.95 
918 

9164 K(14, 9) = -1.16827: K(15, 9) --1.13075: K(16, 9) --1.09338: K(17 

, 9) = -1.05631: K(18, 9) = -1.01973: K(19, 9) --.98381: K(20, 9) = -.94 
871 
9166 K(14, 10) --1.04144: K(15, 10) --1.0181: K(16, 10) --.99418: K(l 
7, 10) = -.96977: K(18, 10) = -.94496: K(19, 10) --.91988: K(20, 10) = .
89464 
9168 K(14, 11) = -.83223: K(15, 11) --.82516: K(16, 11) --.8172: K(17, 

11) = -.80837: K(18, 11) = -.79868: K(19, 11) --.78816: K(20, 11) --.7 
7686 
9170 K(14, 12) = -.63779: K(15, 12) = -.6408: K(16, 12) --.643: K(17, 1 

2) = -.64436: K(18, 12) = -.64488: K(19, 12) = -.64453: K(20, 12) = -.643 
33 
9172 K(14, 13) = -.43949: K(15, 13) --.44942: K(16, 13) = -.45873: K(17 

, 13) = -.46739: K(18, 13) = -.47538: K(19, 13) = -.48265: K(20, 13) = -. 
48917 
9174 K(14, 14) = -.37824: K(15, 14) --.38977: K(16, 14) = -.40075: K(17 
, 14) = -.41116: K(18, 14) = -.42095: K(19, 14) • -.43008: K(20, 14) --. 
43854 
9176 K(14, 15) = -.22535: K(15, 15) = -.23996: K(16, 15) = -.25422: X(17 
, 15) = -.26808: K(18, 15) = -.2815: K(19, 15) = -.29443: K(20, 15) --.3 
0685 
9178 K(14, 16) = -.05803: K(15, 16) = -.07476: K(16, 16) = -.09132: K(17 
, 16) = -.10769: K(18, 16) = -.12381: K(19, 16) = -.13964: K(20, 16) = -. 
15516 
9180 K(14, 17) = .01824: K(15, 17) -.00092: K(16, 17) --.01631: K(17, 
17) = -.03344: K(18, 17) = -.0504: K(19, 17) = -.06718: K(20, 17) --.083 
71 
9182 K(14, 18) = .31307: K(15, 18) -.29535: K(16, 18) = .2774: K(17, 18 
) = .25925: K(18, 18) = .24094: K(19, 18) = .2225: K(20, 18) = .20397 
9184 K(14, 19) = .70512: K(15, 19) = .6905: K(16, 19) = .67532: K(17, 19 
) = .65959: K(18, 19) = .64335: K(19, 19) -.62662: K(20, 19) -.60944 
9186 K(14, 20) -1.33665: K(15, 20) -1.3333: K(16, 20) = 1.329: K(17, 2 

0) = 1.32376: K(18, 20) = 1.3176: K(19, 20) = 1.31054: K(20, 20) = 1.3025 
9 
9188 K(14, 21) = 1.93836: K(15, 21) = 1.95083: K(16, 21) = 1.96213: K(17 
, 21) = 1.97227: K(18, 21) = 1.98124: K(19, 21) = 1.98906: K(20, 21) = 1. 
99573 
9190 K(14, 22) = 2.12768: K(15, 22) -2.14591: K(16, 22) -2.16293: X(17 
, 22) = 2.17873: K(18, 22)'= 2.19332: K(19, 22) = 2.2067: K(20, 22) = 2.2 
1888 
9192 K(14, 23) = 2.52102: K(15, 23) = 2.55222: K(16, 23) -2.58214: K(17 
, 23) = 2.61076: K(18, 23) = 2.6381: K(19, 23) = 2.66413: K(20, 23) = 2.6 
8888 

9194 K(14, 24) -2.70556: K(15, 24) -2.74325: K(16, 24) -2.77964: K(17 
, 24) = 2.81472: K(18, 24) = 2.84848: K(19, 24) = 2.88091: K(20, 24) '= 2. 
91202 
9196 X(14, 25) = 3.27134: K(15, 25) = 3.33035: K(16, 25) = 3.38804: K(17 
, 25) = 3.44438: K(18, 25) = 3.49935: K(19, 25) = 3.55295: K(20, 25) = 3. 
60517 

B-7 


9198 K(14, 26) = 3.82798: K(15, 26) = 3.90973: K(16, 26) = 3.99016: K(l7 

, 26) = 4.06926: K(18, 26) = 4.147: K(19, 26) = 4.22336: K(20, 26) = 4.29 

832 
9200 K(14, 27) = 4.55304: K(15, 27) = 4.66651: K(16, 27) s 4.77875: K(17 
, 27) = 4.88971: K(18, 27) = 4.99937: K(19, 27) -5.10768: K(20, 27) = 
21461 
9202 K(14, 28) = 5.09505: K(15, 28) = 5.23353: K(16, 28) = 5.37087: K(17 
, 28) = 5.50701: K(18, 28) = 5.6419: K(19, 28) = 5.77549: K(20, 28) = 5.9 
0776 
9204 K(14, 29) = 5.63252: K(15, 29) -5.79673: K(16, 29) -5.9599: K(17, 

29) = 6.12196: K(18, 29) = 6.28285: K(19, 29) -6.44251: K(20, 29) = 6.6 

009 
9206 K(14, 30) = 6.86661: K(15, 30) -7.09277: K(16, 30) = 7.31818: X(17 

, 30) = 7.54272: X(18, 30) = 7.76632: K(19, 30) = 7.98888: K(20, 30) = 8. 
21034 
9208 K(21, 0) = -.95234: K(22, 0) --.90908: K(23, 0) --.86956: K(24, 0 
)= -.83333: K(25, 0) = -.8: K(26, 0) --.76923: K(27, 0) --.74074 

9210 K(21, 1) = -.95215: K(22, 1) --.90899: K(23, 1) = -.86952: K(24, 1 
) --.83331: K(25, 1) --.79999: K(26, 1) --.76923: K(27, 1) --.74074 
9212 K(21, 2) = -.95188: K(22, 2) --.90885: K(23, 2) = -.86945: K(24, 2

) = -.83328: K(25, 2) = -.79998: K(26, 2) = -.76922: K(27, 2) = -.74074 

9214 K(21, 3) = -.95131: K(22, 3) --.90854: K(23, 3) = -.86929: K(24, 3 
) = -.8332: K(25, 3) = -.79994: K(26, 3) --.7692: K(27, 3) = -.74073 
9216 K(21, 4) = -.94945: K(22, 4) • -.90742: K(23, 4) = -.86863: K(24, 4 
) = -.83283: K(25, 4) = -.79973: K(26, 4) = -.76909: K(27, 4) • -.74067 
9218 K(21, 5) = -.94607: K(22, 5) = -.90521: K(23, 5) = -.86723: K(24, 5 
)= -.83196; K(25, 5) = -.79921: K(26, 5) --.76878: K(27, 5) --.74049 
9220 K(21, 6) = -.93878: K(22, 6) --.90009: K(23, 6) = -.86371: K(24, 6 
) = -.82959: K(25, 6) --.79765: K(26, 6) --.76779: K(27, 6) = -.73987 
9222 X(21, 7) = -.93495: K(22, 7) --.89728: K(23, 7) = -.86169: K(24 ; 
) = -.82817: K(25, 7) = -.79667: K(26, 7) = -.76712: K(27, 7) = -.7394 
9224 K(21, 8) = -.92295: K(22, 8) --.88814: K(23, 8) = -.85486:' K(24, 8 
) --.82315: K(25, 8) = -.79306: K(26, 8) = -.76456: K(27, 8) = -.73765 
9226 K(21, 9) = -.91458: K(22, 9) = -.88156: K(23, 9) = -.84976: K(24, 9 
) --.81927: K(25, 9) = -.79015: K(26, 9) = -.76242: K(27, 9) --.7361 
9228 K(21, 10) = -.86938: K(22, 10) s -.84422: K(23, 10) = -.81929: K(24 
, 10) = -.79472: K(25, 10) = -.77062: K(26, 10) = -.74709: K(27, 10) --. 
72422 
9230 K(21, 11) = -.76482: K(22, 11) = -.75211: K(23, 11) --.7388: K(24, 

11) = -.72495: K(25, 11) = -.71067: K(26, 11) = -.69602: K(27, 11) = -.6 
8111 
9232 K(21, 12) --.64125: K(22, 12) = -.63833: K(23, 12) = -.63456: K(24 

, 12) = -.62999: K(25, 12) = -.62463: K(26, 12) = -.61854: K(27, 12) = -. 

61176 
9234 K(21, 13) = -.49494: K(22, 13) = -.49991: K(23, 13) --.50409: K(24 

, 13) = -.50744: K(25, 13) = -.50999: K(26, 13) = -.51171: K(27, 13) = -. 

51263 
9236 K(21, 14) = -.44628: K(22, 14) = -.45329: K(23, 14) = -.45953: K(24 

,14) = -.46499: K(25, 14).= -.46966: K(26, 14) = -.47353: K(27, 14) = -. 

4766 
9238 K(21, 15) = -.31872: K(22, 15) = -.32999: K(23, 15) --.34063: K(24 
, 15) = -.35062: K(25, 15) = -.35992: K(26, 15) = -.36852: K(27, 15) --. 
3764 
9240 K(21, 16) = -.1703: K(22, 16) = -.18504: K(23, 16) --.19933: K(24, 

16) = -.21313: K(25, 16) = -.22642: K(26, 16) = -.23915: K(27, 16) = -·2 
5129 
9242 K(21, 17) = -.09997: K(22, 17) = -.1159: K(23, 17) = -.13148: K(~" 


17) = -.14665: K(25, 17) = -.16138: K(26, 17) = -.17564: K(27, 17) = -.1 
8939 

9244 K(21, 18) = .1854: K(22, 18) = .16682: K(23, 18) = .14827: K(24, 18 

8-8 


) = .12979: K(25, 18) = .11143: K(26, 18) = .09323: K(27, 18) = .07523 

9246 K(21, 19) = .59183: K(22, 19) = .57383: K(23, 19) = .55549: K(24, 1 

9) = .53683: K(25, 19) = .51789: K(26, 19) = .49872: K(27, 19) = .47934 

9248 K(21, 20) = 1.29377: K(22, 20) = 1.28412: K(23, 20) = 1.27365: K(24 
, 20) • 1.2624: K(25, 20) = 1.25039: K(26, 20) = 1.23766: K(27, 20) = 1.2 
2422 
9250 K(21, 21) -2.00128: K(22, 21) • 2.0057: K(23, 21) • 2.00903: K(24, 

21) -2.01128: K(25, 21) -2.01247: K(26, 21) -2.01263: K(27, 21) • 2.0 
1177 
9252 K(21, 22) = 2.22986: K(22, 22) • 2.23967: K(23, 22) -2.24831: K(24 
, 22) -2.25581: K(25, 22) = 2.26217: K(26, 22) -2.26743: K(27, 22) -2. 
2716 
9254 K(21, 23) = 2.71234: K(22, 23) -2.73451: K(23, 23) = 2.75541: K(24 
, 23) = 2.77506: K(25, 23) = 2.79345: K(26, 23) = 2.81062: K(27, 23) = 2. 
82658 
9256 K(21, 24) = 2.94181: K(22, 24) • 2.97028: K(23, 24) = 2.99744: K(24 
, 24) = 3.0233: K(25, 24) = 3.04787: K(26, 24) -3.07116: K(27, 24) = 3.0 
932 
9258 K(21, 25) = 3.656: K(22, 25) • 3.70543: K(23, 25) -3.75347: K(24, 
25) = 3.80013: K(25, 25) = 3.8454: K(26, 25) -3.8893: K(27, 25) • 3.9318 
3 
9260 K(21, 26) = 4.37186: K(22, 26) = 4.44398: K(23, 26) = 4.51467: K(24 
, 26) = 4.58393: K(25, 26) = 4.65176: K(26, 26) -4.71815: K(27, 26) = 4. 
78313 
9262 K(21, 27) = 5.32014: K(22, 27) -5.42426: K(23, 27) = 5.52694: K(24 

, 27) = 5.62818: K(25, 27) = 5.72796: K(26, 27) -5.82629: K(27, 27) -5. 

92316 
9264 K(21, 28) = 6.03865: K(22, 28) -6.16816: K(23, 28) • 6.29626: K(24 
, 28) -6.42292: K(25, 28) =·6.54814: K(26, 28) -6.67191: K(27, 28) -6. 
79421 
9266 K(21, 29) = 6.75798: K(22, 29) -6.9137: K(23, 29) -7.06804: K(24, 

29) = 7.22098: K(25, 29) = 7.3725: K(26, 29) = 7.52258: K(27, 29) • 7.67 
121 
9268 K(21, 30) = 8.43064: K(22, 30) • 8.64971: K(23, 30) = 8.86753: K(24 
, 30) -9.08403: K(25, 30) = 9.2992: K(26, 30) = 9.51301: K(27, 30) = 9.7 
25429 
9270 K(28, 0) = -.71429: K(29, 0) --.68966: K(30, 0) = -.66667: K(31, 0

) = -.64516: K(32, 0) = -.625: K(33, 0) = -.60606: K(34, 0) • -.58824 

9272 X(28, 1) = -.71429: K(29, 1) = -.68966: K(30, 1) = -.66667: K(31, 1) = -.64516: K(32, 1) = -.625: X(33, 1) = -.60606: K(34, 1) --.58824 
9274 K(28, 2) = -.71428: K(29, 2) = -.68965: X(30, 2) = -.66667: X(31, 2 
) = -.64516: K(32, 2) --.625: K(33, 2) = -.60606: K(34, 2) --.58824 
9276 K(28, 3) = -.71428: K(29, 3) = -.68965: K(30, 3) = -.66667: K(31, 3) = -.64516: K(32, 3) = -.625: K(33, 3) = -.60606: K(34, 3) --.58824 
9278 K(28, 4) = -.71425: K(29, 4) = -.68964: K(30, 4) = -.66666: K(31, 4 
)= -.64516: K(32, 4) = -.625: K(33, 4) --.60606: K(34, 4) • -.58824 
9280 X(28, 5) = -.71415: K(29, 5) --.68959: K(30, 5) = -.66663: K(31, 5 
) --.64514: K(32, 5) = -.62499: K(33, 5) --.60606: K(34, 5) = -.58823 
9282 K(28, 6) --.71377: K(29, 6) • -.68935: K(30, 6) • -.66649: K(31, 6 
)= -.64507: K(32, 6) --.62495: K(33, 6) --.60603: K(34, 6) --.58822 
9284 X(28, 7) = -.71348: K(29, 7) = -.68917: X(30, 7) --.66638: X(31, 7 
) --.645: K(32, 7) = -.62491: X(33, 7) --.60601: X(34, 7) --.58821 
9286 X(28, 8) --.71227: K(29, 8) --.68836: K(30, 8) --.66585: K(31, 8 
) • -.64465: K(32, 8) --.62469: X(33, 8) --.60587: K(34, 8) • -.58812 
9288 X(28, 9) • -.71116: K(29, 9) • -.68759: K(30, 9) • -.66532: X(31, 9) = -.64429: K(32, 9) = -.62445: K(33, 9) --.60572: X(34, 9) • -.58802 
9290 K(28, 10) = -.70209: K(29, 10) = -.68075: K(30, 10) • -.66023: K(31 
, 10) = -.64056: K(32, 10) = -.62175: X(33, 10) = -.60379: K(34, 10) = -. 
58666 
9292 K(28, 11) = -.66603: K(29, 11) = -.65086: K(30, 11) = -.63569: KC31 

8-9 


, 11) = -.6206: K(32, 11) = -.60567: K(33, 11) = -.59096: K(34, 11) = -.5 
7652 
9294 K(28, 12) = -.60434: K(29, 12) = -.59634: K(30, 12) = -.58783: K(31 

., 12) --.57887: K(32, 12) = -.56953: K(33, 12) = -.55989: K(34, 12) --. 
55 
9296 K(28, 13) = -.51276: K(29, 13) = -.51212: K(30, 13) = -.5107301~ 
~( 
31, 13) = -.50863: K(32, 13) = -.5058501: K(33, 13) --.50244: K(34, 13) 

= -.49844 

9298 K(28, 14) --.47888: K(29, 14) --.48037: K(30, 14) = -.48109: K(31 
, 14) = -.48107: K(32, 14) = -.48033: X(33, 14) --.4789: X(34, 14) --.4 
7682 

9300 K(28, 15) = -.38353: X(29, 15) = -.38991: X(30, 15) = -.39554: K(31 

, 15) --.40041: K(32, 15) = -.40454: X(33, 15)= -.40792: X(34, 15) --. 
41058 
9302 K(28, 16) = -.26282: K(29, 16) = -.27372: X(30, 16) = -.28395: K(31 

, 16) = -.29351: K(32, 16) = -.30238: K(33, 16) = -.31055: K(34, 16) = -. 

31802 
9304 K(28, 17) = -.20259: K(29, 17) • -.21523: X(30, 17) --.22726: K(31 
,17) • -.23868: K(32,17) = -.24946: K(33, 17) --.25958: X(34, 17) --. 
26904 
9306 K(28, 18) = .05746: K(29, 18) = .03997: X(30, 18) = .02279: X(31, 1 

8) = .00596: K(32, 18) = -.0105: X(33, 18) = -.02654: K(34, 18) • -.04215 

9308 K(28, 19) = .4598: K(29, 19) -.44015: X(30, 19) = .4204: K(31, 19) 

= .40061: X(32, 19) = .38081: K(33, 19) = .36104: X(34, 19) -.34133 
9310 K(28, 20) = 1.21013: K(29, 20) = 1.19539: K(30, 20) -1.18006: K(31 
, 20) = 1.16416: K(32, 20) = 1.14772: K(33, 20) -1.13078: X(34, 20) = 1. 
11337 
9312 K(28, 21) = 2.00992: K(29, 21) • 2.0071: K(30, 21) -2.00335: K(31, 

21) = 1.99869: K(32, 21) = 1.99314: X(33, 21) = 1.98674: X(34, 21) -1.9 

7951 
9314 K(28, 22) = 2.2747: K(29, 22) = 2.27676: K(30, 22) = 2.2778: KC' 


22) = 2.27785: K(32, 22) = 2.27693: X(33, 22) = 2.27506: X(34, 22) = ~.~7 
229 
9316 K(28, 23) = 2.84134: K(29, 23) = 2.85492: K(30, 23) -2.86735: K(31 

, 23) -2.87865: K(32, 23) = 2.88884: K(33, 23) = 2.89795: X(34, 23) = 2. 

90599 
9318 K(28, 24) = 3.11399: K(29, 24) = 3.13356: K(30, 24) = 3.15193: K(31 

, 24) = 3.16911: K(32, 24) = 3.18512: K(33, 24) = 3.2: K(34, 24) = 3.2137 

5 
9320 K(28, 25) = 3.97301: K(29, 25) = 4.01286: K(30, 25) = 4.05138: K(31 

, 25) = 4.08859: K(32, 25) = 4.12452: K(33, 25) = 4.15917: K(34, 25) = 4. 

19257 

9322 K(28, 26) = 4.84669: K(29, 26) = 4.90884: K(30, 26) = 4.96959: K(31 

, 26) = 5.02897: K(32, 26) = 5.08697: K(33, 26) = 5.14362: K(34, 26) = 5. 

19892 
9324 K(28, 27) = 6.01858: K(29, 27) = 6.11254: K(30, 27) = 6.20506: K(31 


, 27) = 6.29613: K(32, 27) = 6.38578: K(33, 27) -6.47401: K(34, 27) = 6. 
56084 
9326 K(28, 28) = 6.91505:. K(29, 28) -7.03443: K(30, 28) -7.15235: K(31 
, 28) • 7.26881: K(32, 28) = 7.38382: X(33, 28) = 7.49739: K(34, 28) = 7. 
60953 


9328 K(28, 29) = 7.81839: K(29, 29) -7.96411: K(30, 29) -8.10836: K(31 
, 29) • 8.25115: K(32, 29) = 8.39248: K(33, 29) = 8.53236: K(34, 29) = 8. 
67079 
9330 K(28, 30) • 9.93643: X(29, 30) = 10.14602: K(30, 30) • 10.35418: K( 
31, 30) • 10.5609: K(32, 30) = 10.76618: K(33, 30) -10.97001: K(34, , 


= 11.17239 

9332 K(35, 0) = -.57143: K(36, 0) = -.55556: K(37, 0) = -.54054: K(38, 0

) = -.52632: K(39, 0) = -.51282: K(40, 0) = -.5: K(41, 0) = -.4878 

9334 K(35, 1) = -.57143: K(36, 1) = -.55556: K(37, 1) = -.54054: K(38, 1 

8-10 


) = -.52632: K(39, 1) = -.51282: K(40, 1) = -.5: K(41, 1) = -.4878 

9336 K(35, 2) = -.57143: K(36, 2) = -.55556: K(37, 2) = -.54054: K(38, 2 

)= -.52632: K(39, 2) = -.51282: K(40, 2) = -.5: K(41, 2) = -.4878 

9338 K(35, 3) = -.57143: K(36, 3) = -.55556: K(37, 3) = -.54054: K(38, 3

) = -.52632: K(39, 3) = -.51282: K(40, 3) = -.5: K(41, 3) --.4878 

9340 K(35, 4) = -.57143: K(36, 4) --.55556: K(37, 4) --.54054: K(38, 4 
) --.52632: X(39, 4) = -.51282: X(40, 4) = -.5: X(41, 4) --.4878 
9342 K(35, 5) = -.57143: X(36, 5) --.55556: X(37, 5) --.54054: X(38, 5) = -.52632: X(39, 5) --.51282: K(40, 5) --.5: X(41, 5) --.4878 
9344 K(35, 6) = -.57142: X(36, 6) --.55555: X(37, 6) = -.54054: X(38, 6) = -.52631: X(39, 6) = -.51282: X(40, 6) --.5: X(41, 6) --.4878 
9346 K(35, 7) = -.57141: X(36, 7) --.55555: X(37, 7) --.54054: X(38, 7 
) --.52631: X(39, 7) = -.51282: X(40, 7) --.5: X(41, 7) • -.4878 
9348 K(35, 8) = -.57136: X(36, 8) --.55552: X(37, 8) --.54052: X(38, 8 
)= -.5263: X(39, 8) = -.51281: X(40, 8) = -.5: X(41, 8) --.4878 
9350 X(35, 9) = -.5713: X(36, 9) • -.55548: X(37, 9) --.5405: X(38, 9) = -.52629: X(39, 9) = -.51281: X(40, 9) --.49999: X(41, 9) = -.4878 
9352 K(35, 10) = -.57035: X(36, 10) • -.55483: X(37, 10) = -.54006: X(38 
, 10) = -.526: X(39, 10) = -.5126101: X(40, 10) --.49986: X(41, 10) = -. 
48772 
9354 K(35, 11) = -.56242: X(36, 11) --.54867: X(37, 11) = -.53533: X(38 
, 11) = -.5224001: X(39, 11) = -.5099: X(40, 11) --.49784: X(41, 11) =.
48622 
9356 X(35, 12) = -.53993: X(36, 12) --.52975: X(37, 12) = -.51952: X(38 
, 12) --.50929: X(39, 12) = -.49911: X(40, 12) = -.48902: X(41, 12) --. 
47906 
9358 K(35, 13) = -.49391: X(36, 13) = -.48888: X(37, 13) --.48342: X(38 
, 13) --.47758: X(39, 13) = -.47141: X(40, 13) --.46496: X(41, 13) --. 
45828 
9360 K(35, 14) = -.47413: X(36, 14) = -.47088: X(37, 14) --.46711: X(38 
, 14) = -.46286: X(39, 14) = -.45819: K(40, 14) =-.45314: X(41, 14) --. 
44777 
9362 K(35, 15) = -.41253: X(36, 15) = -.41381: X(37, 15) = -.41442: X(38 

, 15) = -.41441: X(39, 15) = -.41381: X(40, 15) = -.41265: X(41, 15) = -. 

41097 
9364 X(35, 16) = -.32479: X(36, 16) = -.33085: X(37, 16) = -.33623: X(38 
, 16) = -.34092: X(39, 16) = -.34494: X(40, 16) • -.34831: X(41, 16) --. 
35105 
9366 K(35, 17) = -.27782: X(36, 17) = -.28592: X(37, 17) = -.29335: X(38 

, 17) = -.3001: X(39, 17) = -.30617: X(40, 17) = -.31159: X(41, 17) = -.3 

1635 
9368 X(35, 18) = -.0573: X(36, 18) = -7.195001E-02: X(37, 18) = -.0861: 
X(38, 18) = -.09972: X(39, 18) = -.11279: X(40, 18) --.1253: X(41, 18) = 


-.13725 
9370 X(35, 19) = .32171: K(36, 19) = .30223: K(37, 19) = .2829: K(38, 19 
)= .26376: K(39, 19) = .24484: K(40, 19) -.22617: K(41, 19) -.20777 
9372 K(35, 20) = 1.09552: K(36, 20) = 1.07726: K(37, 20) = 1.05863: K(38 
, 20) = 1.03965: K(39, 20} = 1.02036: X(40, 20) -1.00079: X(41, 20) -.9 
8096 
9374 K(35, 21) = 1.97147: K(36, 21) = 1.96266: X(37, 21) • 1.95311: K(38 
, 21) -1.94283: K(39, 21) = 1.93186: X(40, 21) -1.92023: X(41, 21) = 1. 
90796 
9376 K(35, 22) = 2.26862: K(36, 22) = 2.26409: X(37, 22) -2.25872: K(38 
, 22) -2.25254: K(39, 22) = 2.24558: X(40, 22) -2.23786: X(41, 22) -2. 
2294 
9378 K(35, 23) = 2.91299: X(36, 23) = 2.91898: X(37, 23) -2.92397: X(38 
, 23) = 2.92799: K(39, 23) = 2.93107: X(40, 23) = 2.93324: X(41, 23) = 2. 

9345 
9380 K(35, 24) = 3.22641: K(36, 24) = 3.238: X(37, 24) = 3.24853: X(38, 


24) = 3.25803: K(39, 24) = 3.26653: X(40, 24) = 3.27404: X(41, 24) = 3.28 

B-11 


9382 K(35, 25) = 4.22473: K(36, 25) = 4.25569: K(37, 25) = 4.28545: K(38 

, 25) = 4.31403: K(39, 25) = 4.34147: K(40, 25) = 4.36777: K(41, 25) = 4. 
39296 
9384 K(35, 26) = 5.25291: K(36, 26) = 5.30559: K(37, 26) = 5.35698: K(18 

, 26) = 5.40711: K(39, 26) = 5.45598: K(40, 26) = 5.50362: K(41, 26) = 
55005 
9386 K(35, 27) = 6.64627: K(36, 27) = 6.73032: K(37, 27) -6.81301: K(38 
, 27) = 6.89435: K(39, 27) = 6.97435: K(40, 27) = 7.05304: X(41, 27) = 7. 
13043 
9388 K(35, 28) = 7.72024: K(36, 28) • 7.82954: K(37, 28) = 7.93744: K(38 

, 28) = 8.04395: K(39, 28) = 8.1491: X(40, 28) = 8.25289: X(41, 28) = 8.3 
5534 
9390 K(35, 29) = 8.80779: K(36, 29) = 8.94335: X(37, 29) = 9.077501: K(3 

8, 29) = 9.21023: K(39, 29) = 9.341581: K(40, 29) = 9.471541: K(41, 29) = 

9.60013 
9392 K(35, 30) = 11.37334: X(36, 30) = 11.57284: X(37, 30) = 11.77092: K 
(38, 30) = 11.96757: K(39, 30) = 12.1628: X(40, 30) = 12.35663: K(41, 30) 

= 12.54906 

9394 FOR G = 42 TO83 
9396 FOR P = 0 TO 30 
9398 K(G, P) = -K(G-41,30-P) 
9400 NEXT P 
9402 NEXT G 
9404 LOCATE 20, 20 
9406 COLOR 12 

9408 INPUT "PRESS ENTER .TO PROCEED ............... , ERM


" 

9410 CHAIN "FANINP" 
9412 END 

10000 REM ************************************************************ 
10002 REM QUIT PROGRAM 
10004 REM ************************************************************ 
10006 OPEN "QUIT" FOR OUTPUT AS 4t3 
10008 CLS 
10010 COLOR 12 

10012 FOR I = 1 TO 10: PRINT : NEXT I 

10014 PRINT " PROGRAM TERMINATED BY USER" 
10016 COLOR 7 
10020 SYSTEM 
11111 END 

8-12 



1 REM 
2 REM FFFFFFFF AAA NNN NNN 111111111 NNN NNN PPPPPPPP 
3 REM FFF AAAAA NNNN NNN III NNNN NNN PPP PPP 
4 REM FFF AAA AAA NNNNNN NNN III NNNNNN NNN PPP PPP 
5 REM FFFFFF AAA AAA NNN NNN NNN III NNN NNN NNN PPPPPPPP 
6 REM FFF AAAAAAAAAAAA NNN NNNNNN III NNN NNNNN PPP 
7 REM FFF AAA AAA NNN NNNN III NNN NNNN PPP 
8 REM FFF AAA AAA NNN NNN 111111111 NNN NNN PPP 
9 REM 
10 REM 
11 REM Portions (c) Copyright 

12 REM Microsoft corporation 

13 REM 1982-1988 

14 REM All Rights Reserved 

15 REM 

16 REM Alluvial Fan -Michael Baker Jr., Inc. -JULY 1990 

17 REM 

18 REM 

19 COMMON SHARED PC), K(), KG(), KO() 

20 REM 

30 ON KEY(l) GOSUB 10000 

31 KEY(l) ON 

35 REM 

74 REM 

99 REM 

100 REM ************************************************************ 

102 REM INPUT DATA 
104 REM ************************************************************ 

105 REM 

106 CLS 

107 COLOR 11: PRINT "Press Fl and then press ENTER to exit": COLOR 2 

108 PRINT : PRINT 

109 PRINT " ENTER THE NAME OF THE ALLUVIAL FAN " 

110 INPUT" ", B$ 

114 CLS : MOLT = 0 

115 COLOR 11: PRINT "Press Fl and then press ENTER to exit": COLOR 2 

116 FOR I = 1 TO 4: PRINT: NEXT I 

8-13 



120 REM ************************************************************ 

122 REM INPUT MULTIPLE-CHANNEL DATA 
124 REM ********************************************************** 
126 REM 
128 PRINT "DO YOU WISH TO COMPUTE ZONE BOUNDARIES" 
130 INPUT "FOR MULTIPLE CHANNELS (YIN)"; MULT$ 
132 IF INSTR(MULT$, "Y") = 0 AND INSTR(MULT$, "y") = 0 THEN GOTO 179 
134 MULT = 1 
136 CLS : COLOR 11: PRINT "Press Fl and then press ENTER to exit": COLOR 
2 
137 FOR I = 1 TO 4: PRINT: NEXT I 
138 INPUT "ENTER SLOPE OF ALLUVIAL FAN ", SLOPE 
140 CLS : COLOR 11: PRINT "Press Fl and then press ENTER to exit": COLOR 
2 
141 FOR I = 1 TO 4: PRINT NEXT I 
142 IF SLOPE> 1 THEN PRINT "SLOPE ("; SLOPE; II) IS TOO LARGE" ELSE GOTO 
146 
144 GOTO 148 
146 IF SLOPE < .000001 THEN PRINT "SLOPE (II; SLOPE; II) IS TOO SMALL" ELSE 
GOTO 152 
148 FOR I = 1 TO 4: PRINT NEXT I 
150 GOTO 138 
152 CLS : COLOR 11: PRINT "Press Fl and then press ENTER to exit": COLOR 

2 
153 FOR I = 1 TO 4: PRINT: NEXT I 
154 PRINT "SLOPE ="; SLOPE . 


156 FOR I =1 TO 4: PRINT: NEXT I 

158 INPUT "ENTER ROUGHNESS COEFFICIENT (N-VALUE) ", NVALUE 
160 CLS : COLOR 11: PRINT "Press Fl and then press ENTER to exit": COu..~. 
2 
161 FOR I = 1 TO 4: PRINT: NEXT I 
162 IF NVALUE > 1 THEN PRINT liN-VALUE (II; NVALUE; ") IS TOO LARGE" ELSE G 
OTO 166 
·164 GOTO 168 
166 IF NVALUE < .001 THEN PRINT liN-VALUE (II; NVALUE; II) IS TOO SMALL" ELS 
E GOTO 172 
168 FOR I = 1 TO 4: PRINT: NEXT I 
170 GOTO 158 
172 CLS : COLOR 11: PRINT "Press Fl and then press ENTER to exit": COLOR 

2 
173 FOR I = 1 TO 4: PRINT NEXT I 
174 PRINT "MULTIPLE CHANNEL PARAMETERS :" 


176 PRINT II SLOPE =";SLOPE 

178 PRINT II N-VALUE --II., NVALUE 
179 COLOR 11 
180 IF INSTR(MULT$, "Y") = 0 AND INSTR(MULT$, "y") = 0 THEN CLS 
181 IF INSTR(MULT$, "Y") =0 AND INSTR(MULT$, lIyll) = 0 THEN PRINT "Press 
Fl and then press ENTER to exit" 
182 COLOR 2 
183 FOR I = 1 TO 4: PRINT: NEXT I 


8-14 



·184 REM 
186 REM ************************************************************ 

188 REM INPUT AVULSION FACTOR 
190 REM ************************************************************ 

192 REM 
194 INPUT "ENTER AVULSION FACTOR ", AVUL 
196 IF AVUL = 0 THEN AVUL = 1 


199 REM 
200 REM ************************************************************ 

202 REM CHOOSE OPTION FOR DEFINING FLOOD FREQUENCY CURVE 
204 REM ************************************************************ 

205 REM 
206 CLS : COLOR 11: PRINT "Press F1 and then press ENTER to exit": COLOR 
2 
207 FOR I = 1 TO 4: PRINT: NEXT I 
208 PRINT "YOU MAY DEFINE THE FLOOD FREQENCY CURVE BY:" 
210 PRINT : PRINT 
212 PRINT "(l) •••• ENTERING THE MEAN, STANDARD DEVIATION, AND SKEW COEFFIC 
lENT" 
214 PRINT " OF THE PEARSON TYPE-III DISTRIBUTION" 
216 PRINT 
218 PRINT "(2) •••• ENTERING (AT LEAST THREE) PAIRS OF RETURN INTERVALS AND 

DISCHARGES" 
220 PRINT : PRINT 
222 INPUT "PLEASE ENTER OPTION NUMBER ("lOR 2) ", PDFOPT 
224 IF PDFOPT = 1 THEN GOTO 300 


226 IF PDFOPT = 2 THEN GOTO 400 

228 CLS COLOR 11: PRINT "Press F1 and then press ENTER to exit": COLOR 
2 

229 FOR I =1 TO 4: PRINT: NEXT I 

230 PRINT "SORRY, THERE ARE ONLY TWO OPTIONS." 
232 PRINT PDFOPT; " IS NOT ONE OF THEM." 
234 FOR I = 1 TO 4: PRINT: NEXT I 
236 GOTO 208 


8-15 



299 REM 

300 REM ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••~~ 
302 REM OPTION (1) ENTER STATISTICS 

304 REM ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••~* 
305 REM 

306 CLS : COLOR 11: PRINT "Press Fl and then press ENTER to exit": COLOR 

2: FOR I =1 TO 4: PRINT: NEXT I 
307 INPUT" ENTER MEAN ", MU 

308 PRINT 

309 INPUT n ENTER STANDARD DEVIATION ", SIGMA 

310 PRINT 

311 IF SIGMA < .1 THEN GOTO 330 

312 INPUT" ENTER SKEW COEFFICIENT ", SKEW 

313 IF SKEW> 4.1 THEN GOTO 340 

314 IF SKEW < -4.1 THEN GOTO 350 

318 IF SKEW < 0 THEN SK = 4.1 -SKEW ELSE SK = SKEW 

320 G = INT(10 * (SK + .05» 

322 SKEW =G / 10 

324 IF SKEW> 4.1 THEN SKEW = 4.1 -SKEW 

326 GOTO 700 

330 PRINT "SORRY, STANDARD DEVIATION MUST BE GREATER THAN 0.1" 

332 INPUT "RE-ENTER STANDARD DEVIATION ", SIGMA 

334 GOTO 310 

340 PRINT 

341 PRINT "SORRY, SKEW CANNOT BE GREATER THAN 4.1" 

342 INPUT" RE-ENTER SKEW COEFFICIENT ", SKEW 

343 GOTO 313 

350 PRINT 

351 PRINT "SORRY, SKEW CANNOT BE LESS THAN -4.1" 

352 GOTO 342 

8-16 



399 

,-400 
402 
404 
405 
406 
2 
408 
410 
412 
) 
414 
416 

T: 
418 
420 
422 
423 
424 
426 
428 
430 
432 
434 
436 
438 
440 
442 
443 
444 
446 
448 
450 
452 
454 
456 
T: 
458 
460 
462 
464 
476 
478 
480 
2 
481 
482 
484 
486 
488 
490 
492 
494 
REM 

REM ************************************************************ 

REM OPTION (2) ENTER PAIRS OF DATA 

REM ************************************************************ 

REM 

CLS : COLOR 11: PRINT "Press F1 and then press ENTER to exit": COLOR 

FOR I =1 TO 4: PRINT: NEXT I 

PRINT "HOW MANY PAIRS OF DISCHARGES AND" 

INPUT "RECURRENCE INTERVALS 00 YOU WISH TO ENTER"; NOF: NOF = INT(NOF 

IF NOF >= 3 THEN GOTO 424 
CLS : COLOR 11: PRINT "Press F1 and then press ENTER to exit": PRIN 

PRINT : COLOR 
PRINT "SORRY, 
PRINT NOF: " 
GOTO 408 
REM 
PRINT : PRINT 
DIM RET (NOF), 

2 

YOU MUST ENTER AT LEAST THREE PAIRS OF DATA." 
IS LESS THAN THREE." 

Q(NOF), Y(NOF), KK(NOF), PIN (NOF) 

FORI =1TO NOF 

PRINT "ENTER RECURRENCE INTERVAL NUMBER": I: 
INPUT" ", RET(I): GOSUB 1000 
PRINT : PRINT 
PRINT" ENTERII: RET(I): "-YEAR DISCHARGEII: 
INPUT" ", Q(I): GOSUB 2000: GOSUB 3000 
GOSUB 480 

NEXT 

I = NOF 

INPUT" 00 YOU WISH TO CHANGE ANY DATA (YIN)": CHGDAT$ 
IF INSTR(CHGDAT$, IIYII) = 0 AND INSTR(CHGDAT$, "y") = 0 GOTO 500 
PRINT : PRINT 
PRINT "WHICH PAIR OF DATA ( 1 _II; NOF; II) 00 YOU WISH TO"; 
INPUT" CHANGEII; CHGDAT: CHGDAT -INT(CHGDAT) 
IF (CHGDAT >= 1) AND (CHGDAT <= NOF) THEN GOTO 464 
CLS : COLOR 11: PRINT "Press F1 and then press ENTER to exit": PRIN 
PRINT : COLOR 2 
P~INT 
"SORRY, THERE IS NO DATA PAIR NUMBER II; CHGDAT 

PRINT : PRINT 
GOSUB 482: GOTO 
GOSUB 3066 
GOSUB 480 
GOTO 444 
CLS : COLOR 11: 

FOR K = 1 TO 4: 
PRINT 

II 

PRINT 

FORJ =1TOI 

PRINT USING 
NEXT 
PRINT : PRINT : 
RETURN 

444 

PRINT IIPress F1 and then press ENTER to exit": COLOR 

PRINT: NEXT K 
DATA PAIR RECURRENCE INTERVAL DISCHARGE" 

"##################"; J; RET(J); Q(J) 
PRINT 

8-17 


499 REM 

500 REM •••••••••••••••••••••••••••••••••••*.**.*******••*********,


502 REM FIND SKEW THAT GIVES BEST FIT 

504 REM ••*•••*••***•••**.*••***•••••••**.*********************.**** 

505 REM 

506 LEFT = -4.1: RIGHT = 4.1 

508 IF RIGHT -LEFT < .12 GOTO 528 

510 MID = INT(10 * (RIGHT + LEFT) / 2 + .001) / 10 

512 RMID = MID + .1 


514IFMID <0 THENG = 10 * (4.1 -MID) ELSE G = 10 * MID 

516 GOSUB 600: MIDR = R 

518 IF RMID < 0 THEN G = 10 * (4.1 -RMID) ELSE G = 10 * RMID 

520 GOSUB 600: RMIDR = R 

522 IF RMIDR < MIDR GOTO 526 

524 LEFT = MID: GOTO 508 

526 RIGHT = MID: GOTO 508 

528 IF RMIDR > MIDR THEN SKEW = RMID ELSE SKEW = MID 

530 IF SKEW < 0 THEN G=10 * (4.1 -SKEW) ELSE G = 10 * SKEW 

532 GOSUB 600 

534 RMAX = R: MU = MEAN: SIGMA = STDV 

550 GOTO 700 

600 REM 
602 REM .**********.*******••••••****••******••********••****.****** 

604 REM GIVEN INPUT DATA (OPTION 2), COMPUTE 
606 REM LOG(J) AND CORRESPONDING DEVIATE KK(J) 
608 REM **********************************************************~A 


610 REM 

612 FORJ =1 TO NOF 

614 Y(J) = LOG(Q(J» / LOG(lO) 

616 PIN(J) = 1 / RET(J) 

618 FOR I =1 TO 30 
620 IF PIN(J) < P(I) THEN GOTO 626 
622 N=I: M = N-1 

624 GOTO 628 

626 NEXT I 

628 KK(J) = K(G, M) + (PIN(J) -P(M» * (K(G, N) -K(G, M» / (P(N) 


P(M) ) 
630 NEXT J 

8-18 



632 REM 
634 REM ************************************************************ 

636 REM GIVEN NOF PAIRS OF DATA, Y(J) AND KK(J), COMPUTE 
638 REM MEAN (Y-INTERCEPT), STANDARD DEVIATION (SLOPE), 
640 REM AND CORRELATION COEFICIENT BY METHOD OF LEAST SQUARES 
642 REM ************************************************************ 

644 REM 
646 MK = 0: MY = 0: A = 0: B = 0: C = 0 
648 FOR I = 1 TO NOF 
650 MK = MK + KK(I) 
652 MY = MY + Y(I) 
654 NEXT 
656 MEANK = MK/ NOF 
658 MEANY = MY/ NOF 
660 FOR I = 1 TO NOF . 

662 A=A + (KK(I) -MEANK) ~ 
2 

664 B=B + (Y(I) -MEANY) ~ 
2 

666 C = C + (KK(I) -MEANK) * (Y(I) -MEANY) 

668 NEXT 
670 SIGK = SQR(A / NOF): SIGY = SQR(B / NOF) 


672 R=C/ NOF / SIGK / SIGY 

674 STDV = R * SIGY / SIGK 

676 MEAN = MEANY -STDV * MEANK 

678 RETURN 

700 REM 
702 REM ************************************************************ 

704 REM CHECK VALUE OF 100-YEAR FLOOD DISCHARGE 
706 REM ************************************************************ 

708 REM 
709 IF (SIGMA * K(G, 25) + MO) > 6 THEN GOTO 714 

710 Q100 = 10 ~ 
(SIGMA * K(G, 25) + MO) 

712 IF Q100 < 5000001 THEN GOTO 730 
714 CLS : COLOR 12 
716 FOR I = 1 TO 8: PRINT NEXT I 
718 PRINT " Q100 > 500000 cfs .•• PROGRAM TERMINATED": 
PRINT : PRINT : PRINT 
720 COLOR 15, 0: PRINT" Press ENTER to start over 
": PRINT 
721 INPUT n or press F1 and then ENTER to exit", MJM 
722 COLOR 12 
723 GOTO 100 
730 IF Q100 > 50 THEN GOTQ 800 
732 CLS : COLOR 12 
734 FOR I = 1 TO 8: PRINT NEXT I 
736 PRINT " Q100 < 50 cfs ••••••• PROGRAM TERMINATED": 

PRINT : PRINT : PRINT 
738 GOTO 720 

8-19 



800 REM 

802 REM **********************************************************~ 
, 

804 REM WRITE INPUT TO INPUT FILE 

806 REM ************************************************************ 

808 REM 

812 PRINT #1, B$ 

814 PRINT 11, MOLT, SLOPE, NVALUE 

816 PRINT 11, PDFOPT, AVUL 

818 PRINT 11, NOF 

820 FOR I = 1 TO NOF 

822 PRINT #1, RET(I), Q(I), KK(I) 

824 NEXT I 

826 PRINT #1, MO, SIGMA, SKEW, RMAX 

828 FORI =0TO30 

830 PRINT 11, P(I) 

832 NEXT I 

834 FOR I = 0 TO 30 

836 KO(I) s K(O, I}: PRINT 11, KO(I) 

838 KG(I) = K(G, I): PRINT 11, KG(I) 

840 NEXT I 

842 PRINT #1, Ql00 

850 SYSTEM 

B-20 



1000 REM 
1001 REM ************************************************************ 


1002 REM CHECK RECURRENCE INTERVAL 
1004 REM ************************************************************ 


1005 REM 

1006 IF RET(I) < 1.001 GOTO 1012 

1008 IF RET(I) > 1000 GOTO 1024 

1010 GOTO 1030 

1012 PRINT 

1014 PRINT "SORRY, RECURRENCE INTERVAL CANNOT BE LESS THAN 1.001 YEAR" 

1016 PRINT 

1018 PRINT "RE-ENTER RECURRENCE INTERVAL": 

1020 INPUT" ", RET(I) 

1022 GOTO 1000 

1024 PRINT 

1026 PRINT "SORRY, RECURRENCE INTERVAL CANNOT BE GREATER THAN 1000 YEARS" 

1028 GOTO 1016 

1030 IF I = 1 THEN RETURN 

1032 FOR J = 1TOI -1 

1034 IF RET (I) = RET(J) GOTO 1040 

1036 NEXT J 

1038 RETURN 

1040 PRINT 

1042 PRINT "SORRY, EACH RECURRENCE INTERVAL MAY BE ENTERED ONLY ONCE" 

1044 PRINT RET(I): "IS ALSO RECURRENCE INTERVAL NUMBER ": J 

1046 GOTO 1016 

2000 REM 
2001 REM ************************************************************ 


2002 REM CHECK THAT DISCHARGE IS GREATER THAN ZERO 
2004 REM ************************************************************ 


2005 REM 

2006 IF Q(I) > 0 THEN RETURN 

2008 PRINT 

2010 PRINT "DISCHARGE MUST BE GREATER THAN ZERO" 

2012 PRINT 

2014 PRINT "RE-ENTER": RET(I): "-YEAR DISCHARGE": 

2016 INPUT" It, Q(I) 

2018 GOTO 2000 

B-21 



3000 REM 
3001 REM *********************************************************~ 


3002 REM ORDER DATA BY RECURRENCE INTERVAL 
3003 REM AND CHECK THAT DISCHARGES DO NOT DECREASE 
3004 REM ************************************************************ 

3005 IF I = 1 THEN RETURN 
3006 FOR J = 1TOI -1 
3008 IF RET(I) < RET(J) GOTO 3014 
3010 NEXT J 
3012 GOTO 3024 
3014 RET = RET(I): 0 = 0(1) 

3016 FOR K=I TO J + 1 STEP -1 

3018 RET(K) = RET(K -1): O(K) = O(K -1) 

3020 NEXT K 

3022 RET(J) = RET: O(J) = 0 

3024 IF O(J) < O(J -1)" GOTO 3030 

3025 IF J = I THEN RETURN 

3026IFJ <I ANDO(J) >O(J + 1) GOTO 3030 
3028 RETURN 
3030 CLS : COLOR 11: PRINT "Press F1 and then press ENTER to exit": PRI 
NT : PRINT : COLOR 2 
3032 PRINT "SORRY, DISCHARGE VALUES CANNOT DECREASE WITH INCREASING RECUR 
RENCE INTERVALS" 
3034 PRINT 
3036 IF O(J) < O(J -1) THEN L = J ELSE L = J + 1 
3038 PRINT "THE"; RET(L); "-YEAR DISCHARGE ("; O(L); " CFS ) IS LESS THA 


Nil 
3039 PRINT "THE"; RET(L -1); "-YEAR DISCHARGE ("; O(L -1); " CFS )"
3040 PRINT 
3042 PRINT " DATA PAIR RECURRENCE INTERVAL DISCHARGE" 
3044 PRINT 
3046 PRINT USING "##################"; L -1; RET(L -1); O(L -1) 


3048 PRINT USING "##################"; L; RET(L); O(L) 
3050 PRINT : PRINT 
3052 PRINT "WHICH PAIR OF DATA DO YOU WISH TO CHANGE --" 


3054 PRINT "DATA PAIR NUMBER"; L -1; "OR DATA PAIR NUMBER"; L; "?"; 
3056 INPUT" ", CHGDAT 


3058 IF CHGDAT = L -1 OR CHGDAT = L GOTO 3066 

3060 CLS : COLOR 11: PRINT "Press F1 and then press ENTER to exit": PRI 
NT : PRINT : COLOR 2 
3062 PRINT "SORRY, YOU ONLY HAVE TWO CHOICES. "; CHGDAT; "IS NOT ONE OF T 
HEM." 
3064 GOTO 3040 
3066 IF CHGDAT = I GOTO 3074 
3068 FOR K = CHGDATTOI -1 

3070 RET(K) = RET(K + 1): O(K) = O(K + 1)

3072 NEXT K " 
3074 PRINT 
3076 PRINT "RE-ENTER RECURRENCE INTERVAL NUMBER"; CHGDAT; 
3078 INPUT" ", RET(I): GOSUB 1000 
3080 PRINT 
3082 PRINT "ENTER"; RET(I); "-YEAR DISCHARGE"; 
3084 INPUT" ", O(I) 
3085 GOSUB 2000 
3086 GOTO 3000 

8-22 


9999 REM 
10000 REM ************************************************************ 


10002 
REM QUIT PROGRAM 
'REM

10004 ************************************************************ 

10006 OPEN "QUIT" FOR OUTPUT AS #3 
10008 CLS 
10010 COLOR 12 
10012 FOR I = 1 TO 10: PRINT NEXT I 
10014 PRINT " PROGRAM TERMINATED BY USER" 
10016 COLOR 7 
10020 SYSTEM 
11111 END 

B-23 



1 REM 
2 REM FFFFFFFF AAA NNN NNN RRRRRRRR UUU UUU NNN Nh 
3 REM FFF AAAAA NNNN NNN RRR RRR UUU UUU NNNN NNN 
4 REM FFF AAA AAA NNNNNN NNN RRR RRR UUU UUU NNN NNN NNN 
5 REM FFFFFF AAA AAA NNN NNN NNN RRRRRRRR UUU UUU NNN NNNNNN 
7 REM FFF AAAAAAAAAAAA NNN NNNN RRR RRR UUU UUU NNN NNNN 
8 REM FFF AAA AAA NNN NNN RRR RRR UUUUUUU NNN NNN 
9 REM 
10 REM 

11 REM ************************************************************ 

12 REM SET UP CHECK FOR GRAPHICS CAPABILITIES 
13 REM ************************************************************ 

14 ON ERROR GOTO 17 

15 SCREEN 2: PRESET (1, 1) 

16 GOTO 19 

17 GRCHK = 99 

18 RESUME NEXT 

19 ON ERROR GOTO 0 

20 KEY OFF 

100 REM 

102 REM ************************************************************ 

104 REM GET INPUT FROM INPUT FILE -FAN.IN 
106 REM ************************************************************ 

108 REM 

120 DIM P(30), KG(30), KO(30) 

122 OPEN "FAN.IN" FOR INPUT AS #1 

124 OPEN "FAN.OUT" FOR OUTPUT AS #2 

126 INPUT #1, B$ 

128 A$ = SPACE$(INT«71 -LEN(B$» / 2» + B$ 

130 INPUT #1, MULT, SLOPE, NVALUE 

132 INPUT #1, PDFOPT, AVUL 

134 INPUT #1, NOF 

136 DIM RET (NOF), Q(NOF), KK(NOF) 

138 FORI =1 TONOF 

140 INPUT #1, RET(I) , Q(I), KK(I) 

142 NEXT I 

144 INPUT #1, MU, SIGMA, SKEW, RMAX 

146 FOR I = 0 TO 30 

148 INPUT #1, P(I) 

150 NEXT I 

152 FOR I = 0 TO 30 

154 INPUT #1, KO(I), KG(I) 

156 NEXT I 

158 INPUT #1, Q100 

850 GOSUB 6200 

852 IF GRCHK < 44 THEN GOSUB 8200 

854 IF GRCHK < 44 THEN GOTO 1000 

900 CIS 

904 LOCATE 4, 16 

906 PRINT "SYSTEM NOT COMPATIBLE WITH GRAPHICS SUBROUTINES" 

910 LOCATE 20, 30 

911 COLOR 18 

912 PRINT "PROGRAM RUNNING ••• " 

8-24 


,


"1000 REM 
1002 REM ************************************************************ 
1004 REM ************************************************************ 
1006 REM ******************** MAIN PROGRAM ******************** 
1008 REM ************************************************************ 
1010 REM ************************************************************ 
1012 REM 
1014 REM 
1016 REM ************************************************************ 
1018 REM TRANSFORM RANDOM VARIABLE (Y TO Z): 
1019 MUY = MU: SIGMAY = SIGMA 
1020 GOSUB 2000 
1022 REM ************************************************************ 
1024 REM 
1026 GOSUB 6400 
1028 REM 
1030 REM ************************************.*********************** 
1032 REM ASSIGN DISCHARGES AND CALCULATE ALLUVIAL FAN 
1034 REM WIDTHS FOR DEPTH ZONE BOUNDARIES: 
1035 GOSUB 2200 
1036 REM ************************************************************ 
1038 REM 
1040 REM ************************************************************ 
1042 REM ASSIGN DISCHARGES AND CALCULATE ALLUVIAL FAN 
1044 REM WIDTHS FOR VELOCITY ZONE BOUNDARIES: 
1045 GOSUB 2400 
1046 REM ************************************************************ 
1048 REM 
1050 GOSUB 6600 
1051 IF GRCHK < 44 THEN GOSUB 8400 
1052 REM 
1054 REM ************************************************************ 
1056 REM ************************************************************ 
1058 REM ******************** END OF RUN ******************** 
1060 REM ************************************************************ 
1062 REM ************************************************************ 
1064 REM 
1066 REM 
1068 REM ************************************************************ 
1069 REM OPTION TO VIEW AND/OR PRINT OUTPUT DATA: 
1070 GOSUB 9000 
1072 REM ************************************************************ 
1074 REM 
1098 SYSTEM 

8-25 



1999 REM 
2000 REM *********************************************************i 


2002 REM TRANSFORM RANDOM VARIABLE 
2004 REM ************************************************************ 


2005 REM 
2006 IF SKEW = 0 THEN GOTO 2022 
2008 SHAPE = 4 / SKEW / SKEW 

2010 SCALE = 2 / SKEW / SIGMAY 

2012 TRANS -= MOY -2 * SIGMAY / SKEW 

2014 MUZ = TRANS + SHAPE / (SCALE -.92) 

2016 SIGMAZ = SQR{SHAPE / (SCALE -.92) / (SCALE -.92» 

2018 CNST = EXP{.92 * TRANS) * {SCALE / (SCALE -.92» SHAPE

A 

2020 GOTO 2028 

2022 MUZ = MOY + .92 * SIGMAY * SIGMAY 

2024 SIGMAZ = SIGMAY 

2026 CNST = EXP{.92 * HOY + .42 * SIGMAY * SIGMAY) 

2028 RETURN 

2200 REM 
2202 REM ************************************************************ 


2204 REM SUBROUTINE TO COMPUTE CONTOUR WIDTHS FOR DEPTH ZONES 
2206 REM ************************************************************ 


2208 REM 
2210 DH = INT{ (Q100 / 274) .4 + 1)


A 

2212 DIM PHYSING{DH), PHZSING{DH), HSING{DH), QHSING{DH), WHSING{DH) 
2214 DIM PHYMULT{DH), PHZMULT{DH), HMULT{DH), QHMULT{DH), WHMULT{DH) 

2216H -=.5: NH=0 

2218 IF MOLT = 2 THEN GOSUB 2300 ELSE Q = 274.3902 * H 2.5

A 

2220 IF Q > Q100 THEN GOTO 2262 
2222 IF MOLT = 2 THEN NHMULT = NH ELSE NHSING = NH 
2224 NH = NH + 1 
2226 Y • LOG{Q) / LOG(10)
2228 MU = MOY: SIGMA = SIGMAY: GOSUB 4000 

2230 IF MOLT = 2 THEN PHYMULT{NH) = P ELSE PHYSING{NH) = P 

2232 MU = MUZ: SIGMA = SIGMAZ: GOSUB 4000 

2234 IF MOLT = 2 THEN PHZMULT{NH) = P ELSE PHZSING{NH) = P 

2236 IF MOLT = 2 THEN GOTO 2244 

2238 HSING{NH) = H: QHSING{NH) = INT{{INT{Q * 10) +5) / 10) 

2240 WHSING{NH) = AVUL * CNST * PHZSING{NH) * 940.8059 

2242 GOTO 2248 

2244 HMULT{NH) = H: QHMULT{NH) = INT{{INT{Q * 10) +5) / 10) 

2246 WHMULT{NH) = AVUL * CNST * PHZMOLT{NH) * 3575.0624# 
2248 IF MOLT = 2 THEN SORM = 35.750624# ELSE SORM = 9.408059 

2250 IF MOLT = 2 THEN W = WHMULT (NH) ELSE W = WHSING (NH) 

2252 GOSUB 4200 

2254 IF MOLT = 2 THEN WHMULT{NH) = W ELSE WHSING{NH) = W 

2256 H = H +1 
2258 IF MOLT = 2 THEN NHMULT = NH ELSE NHSING -= NH 
2260 GOTO 2218 
2262 MOLT = MOLT + 1 

2264 IF MOLT = 2 THEN GOTO 2216 

2266 RETURN 


2300 

2302 
2304 
2306 
2308 
2310 
2312 
LOPE 
2314 
2315 
2316 
2318 
2320 
2322 
2324 
2326 
2328 
2330 
2332 

REM 
REM ************************************************************ 


REM SUBROUTINE TO COMPUTE Q(H) FOR MULTIPLE CHANNELS 
************************************************************

REM 

REM 
QL z 0: QH-Q100: QG = Q100: HL-0 
HG z 9.220001E-02 * NVALUE A .6 * QG A 

A .6 * QG A .48 / NVALUE A 1.2 
IF QH -QL < .01 THEN GOTO 2330 
IFQL/ QH > .99999 THEN GOTO 2330 
IF (QG = Q100) AND (HG < H) THEN GOTO 
IF QG =Q100 THEN HH =HG 

IF HG > H THEN HH =HG ELSE HL =HG 

IF HG > H THEN QH = QG ELSE QL = QG 

QG =(QH + QL)/ 2 

GOTO 2312 
Qz 2 * Q100: GOTO 2332 

Q = (QH + QL)/ 2 

RETURN 

.36 / SLOPE A .3 + .00143 * S 

2328 

B-27 



2400 REM 
2402 REM *********************************************************J 

2404 REM SUBROUTINE TO COMPUTE CONTOUR WIDTHS FOR VELOCITY ZONES 
2406 REM ************************************************************ 

2408 REM 
2410 DV -INT«Q100 / .12) .2 -2)


A 

2412 DIM PVYSING(DV), PVZSING(DV), VSING(DV), QVSING(DV), WVSING(DV) 
2414 DIM PVYMULT(DV), PVZMULT(DV), VMULT(DV), QVMULT(DV), WVMULT(DV) 
2416 GOTO 2430 
2417 IF MOLT = 4 AND QHMULT(l) < 1 THEN GOTO 2474 


2418 VMAX = .3033 * SLOPE .3 * Q100 A .24 / NVALUE .6

AA 

2420 VMIN = .3033 * SLOPE .3 * QHMULT(l) A .24 / NVALUE .6

AA 

2422 VT = .5 
2424 IF VT > VMIN THEN GOTO 2430 
2426 VT = VT + 1 
2428 GOTO 2424 
2430 NV = 0: IF MOLT = 4 THEN V .. VT ELSE V = 3.5 

2432 IF MOLT = 4 THEN Q = 144.1315 * NVALUE 2.5 * V A (25/ 6) / SLOPE

A 

A 1.25 ELSE Q = .1289 * V5

A 

2434 IF Q > Q100 THEN GOTO 2474 
2436 NV-NV + 1 
2438 Y = LOG(Q) / LOG(10) 

2440 MO .. MOY: SIGMA = SIGMAY: GOSUB 4000 

2442 IF MOLT .. 4 THEN PVYMULT (NV) .. P ELSE PVYSING (NV) = P 
2444 MO -MtTZ: SIGMA = SIGMAZ: GOSUB 4000 
2446 IF MOLT -4 THEN PVZMOLT (NV) -P ELSE PVZSING (NV) = P 
2448 IF MOLT = 4 THEN GOTO 2456 
2450 VSING(NV) = V: QVSING(NV) -INT«INT(Q * 10) + 5)/ 10) 

2452 WVSING(NV) = AVUL * CNST * PVZSING(NV) * 940.8059 

2454 GOTO 2460 

2456 VMULT(NV) = V: QVMULT(NV) = INT«INT(Q * 10) + 5)/ 10) 

2458 WVMULT(NV) = AVUL * CNST * PVZMOLT(NV) * 3575.0624# 

2460 IF MOLT = 4 THEN SORM = 35.750624# ELSE SORM = 9.408059 

2462 IF MOLT = 4 THEN W = WVMULT(NV) ELSE W -WVSING(NV) 
2464 GOSUB 4200 
2466 IF MOLT = 4 THEN WVMULT(NV) = W ELSE WVSING(NV) = W 
2468 V = V + 1 
2470 IF MOLT = 4 THEN NVMULT = NV ELSE NVSING = NV 
2472 GOTO 2432 
2474 MOLT = MOLT + 1 
2476 IF MULT = 4 THEN GOTO 2417 
2478 RETURN 

8-28 



4000 REM 
4002 REM ************************************************************ 


4004 REM SUBROUTINE TO COMPUTE PROBABILITY 
4006 REM GIVEN LOG(Q), MEAN, STANDARD DEVIATION, AND SKEW 
4008 REM ************************************************************ 
4010 REM 
4012 X-(Y -MU) / SIGMA 
4014 IF K > XG(O) THEN GOTO 4018 


4016 P = 1: GOTO 4032 

4018 FOR I = 1 TO 30 
4020 IF X > XG(I) THEN GOTO 4026 
4022 N=I: M=N -1 

4024 GOTO 4030 
4026 NEXT 

4028 P = 0: GOTO 4032 
4030 P = peN) + (P(M) -peN»~ 
* (X -KG(N» / (KG(M) -XG(N» 
4032 RETURN 

4200 REM 
4202 REM ************************************************************ 


4204 REM SUBROUTINE TO ADJUST WIDTH FOR CHANNEL WIDTH > FAN WIDTH 
4206 REM ************************************************************ 


4208 REM 

4209 WI = W: NA=0 

4210 PROB = P 
4212 QW = (W / SORM) A 2.5 

4214 Y = LOG(QW) / LOG(10) 

4216 MU = MUY: SIGMA = SIGMAY: GOSUB 4000 

4218 PYQW = P 

4220 MU = MUZ: SIGMA = SIGMAZ: GOSUB 4000 

4222 PZQW = P 

4224 PRB = AVUL * CNST * SORM/ W * (PROB -PZQW) + PYQW 

4225 IF PRB > .01 THEN NA =W 

4226 IF PRB < .01 THEN WI = W 
4227 WNEW = 100 * PRB * W 

4229 IF ABS(WNEW -W) < 1ORWI-NA < 1 THEN GOTO 4234 
4230 IF ABS(WNEW -W) >= WI-NATHENW = (WI +NA) / 2ELSEW = WNEW 
4232 GOTO 4212 
4234 RETURN 

8-29 



6200 REM 
6202 REM **********************************************************+. 


6204 REM FLOOD FREQUENCY OUTPUT 
6206 REM ***********************************************************


6208 REM 
6210 PRINT t2, CHR$(12) 
6212 PRINT t2, A$ 


6214 FOR I = 1 TO 2: PRINT #2, : NEXT 
6216 PRINT #2, USING" AVULSION FACTOR = #.f###"; A 
WL 
6218 FOR I = 1 TO 4: PRINT #2, : NEXT 


6220 IF PDFOPT = 2 GOTO 6226 

6222 PRINT f2," FLOOD FREQUENCY CURVE DEFINED BY MEAN, STANDARD DEVIA 
TION, AND SKEW" 
6224 GOTO 6242 
6226 PRINT #2," FLOOD FREQUENCY CURVE DEFINED BY LEAST-SQUARES FIT OF 

DATA" 
6228 FOR I = 1 TO 2: PRINT #2, : NEXT 
6230 PRINT #2," RETURN INTERVAL INPUT DISCHARGE BEST FIT DISCHA 
RGE" 
6232 PRINT #2, "(YEARS) (CFS) (CFS) " 
6234 PRINT #2, 
6236 FOR K = 1 TO NOF 
6238 PRINT #2, USING" tt#t tt##ft 
f##.#."; RET(K); Q(K); 10 A (SIGMA * KK(K) + MU) 
6240 NEXT 

6242 FOR I = 1 TO 2: PRINT #2, : NEXT 

6244 PRINT #2, USING II MEAN -,#.###,f 

: MU 

6246 PRINT #2, USING II STANDARD DEVIATION = ##.######" 
; SIGMA 
6248 PRINT #2, USING II SKEW = #'.#"; SKE 
W 
6250 FOR I = 1 TO 4: PRINT #2, : NEXT 
6252 PRINT #2, II SUMMARY OF DISCHARGES:" 
6254 PRINT #2, 
6256 PRINT #2, USING" 10-YEAR DISCHARGE -= #####'11; 1 

o A (SIGMA * KG(20) + MU) 
6258 PRINT #2, USING" 50-YEAR DISCHARGE = ###"#"; 1 
o A (SIGMA * KG(24) + MU) 
6260 PRINT #2, USING" 100-YEAR DISCHARGE = ######"; 1 
o A (SIGMA * KG(25) + MU) 
6262 PRINT #2, USING II 500-YEAR DISCHARGE = ######"; 1 
o A (SIGMA * KG(27) + MU) 
6264 FOR I = 1 TO 4: PRINT #2, : NEXT 
6265 IF SIGMA * SKEW> 2.1 THEN GOTO 6270 

6266 RETURN 
6270 PRINT #2, " STANDARD DEVIATION TIMES SKEW GREATER T . 
HAN 2.1" 
6272 PRINT t2, II WIDTHS CANNOT BE COMPUTEDII 
6274 PRINT #2, 
6276 PRINT #2, " PROGRAM TERMINATED" 
6278 COLOR 10 


B-30 


6279FORI=1TO4:PRINT NEXTI 

6280 PRINT " STANDARD DEVIATION TIMES SKEW GREATER THAN 

2.1" 

6282 PRINT n WIDTHS CANNOT BE COMPUTED" 

6284 PRINT 

6286 PRINT : PRINT : PRINT : COLOR 12 

6290 PRINT n PROGRAM TERMINATED" 

6292 FOR I = 1 TO 4: PRINT: NEXT I 

6294 INPUT" DO YOU WISH TO VIEW THE FLOOD FREQENCY DATA (YIN 

)? ", FFD$ 

6295 IF INSTR(FFD$, "Y") = 0 AND INSTR(FFD$, "y") = 0 THEN GOTO 9900 

6296 HOY = MU: SIGMAY = SIGMA 

6298 COLOR 10: GOTO 9100 

6400 REM 
6402 REM ************************************************************ 


6404 REM TRANSFORMATION OUTPUT 
6406 REM ************************************************************ 


6408 REM 

6412 IF SKEW = 0 GOTO 6420 

6414 AAAA = -.92 * TRANS I (SCALE -.92): BBBB -SCALE I (SCALE -.92)

6416 PRINT '2, USING II STATISTICS AFTER TRANSFORMATION OF Y=LOG(Q) T 

o Z='.'#'#+#.#'" LOG(Q)"; -.92 * TRANS I (SCALE -.92); SCALE I (SCALE .
92) 
6418 GOTO 6424 
6420 AAAA = .92 * SIGMAY * SIGMAY: BBBB = 11 

6422 PRINT #2, USING II STATISTICS AFTER TRANSFORMATION OF Y=LOG(Q) T 

o Z='.'#'#+LOG(Q)II; .92 * SIGMAY * SIGKAY 
6424 PRINT '2, 
6426 PRINT #2, USING MEAN OF Z
II 

; MUZ 
6428 PRINT '2, USING .. STANDARD DEVIATION = ,#.####,#" 
: SIGMAZ 
6430 PRINT #2, USING" SKEW 
: SKEW 
6432 PRINT #2, USING" TRANSFORMATION CONSTANT = #,.#",#,,, 
; CNST 
6434 PRINT #2, CHR$(12) 
6436 RETURN 

B-31 



6600 REM 
6602 REM *********************************************************.... 


6604 REM DEPTH-AND VELOCITY-ZONE OUTPUT 
6606 REM ************************************************************ 
6608 REM 


6610 PRINT '2, B$; SPC(66 -LEN(B$»; "PAGE 2" 
6612 FOR I = 1 TO 4: PRINT #2, : NEXT 
6614 PRINT 12, II SINGLE-CHANNEL REGION II 
6616 PRINT i2, " 


II 

6618 PRINT #2, PRINT #2, 
6620 PRINT #2, II PROBABILITY OF DISC 
HARGE" 
6622 PRINT #2, n BEING EXCEEDED AT 
THEil 
6624 PRINT i2, " ENERGY DEPTH DISCHARGE APEX BY: 


WIDTH" 
6626 PRINT i2, USING II (FT) (FT) (CFS) 

i.I##t (FT)"; BBBB 
6628 PRINT #2, USING II 0 ###. 
#### 0"; 10 A AAAA 
6630 PRINT '2, 

6632 NN = NHSING 

6634 FOR I = 1 TO NN 

6636 DEPTH = 2 * HSING(I) / 3 

6638 PRINT #2, USING II i#.# ,#.# ####### #.""t 
#.###tt ,########"; HSING(I); DEPTH; OHSING(I); PHYSING(I); PHZSING(I 
); WHSING(I) 
6640 NEXT 
6642 FOR I = 1 TO 4: PRINT #2, : NEXT 
6644 PRINT #2, II 

II 

6646 PRINT #2, PRINT #2, 
6648 PRINT '2, II PROBABILITY OF DISC 
HARGE" 
6650 PRINT #2, " BEING EXCEEDED AT 


THEil 
6652 PRINT #2, II VELOCITY DEPTH DISCHARGE APEX BY: 
WIDTH" 
6654 PRINT #2, USING" (FT/SEC) (FT) (CFS) 

#.#### (FT)"; BBBB 
6656 PRINT #2, USING II o ###. 
##t# Q"; 10 A AAAA 
6658 PRINT #2, 

6660 NN = NVSING 

6662 FOR I = 1 TO NN 
6664 DSING = VSING(I) A 2 I 32.16 
6666 PRINT #2, USING" ##., ##.f ####### #.#1##1 
t.ttill #####1###"; VSING(I); DSING; OVSING(I); PVYSING(I); PVZSING(I 
); WVSING{I) 
6668 NEXT 
6670 IF MOLT < 5 GOTO 6742 
6672 PRINT #2, CHR$(12) 
6674 PRINT #2, B$; SPC(66 -LEN(B$»; "PAGE 3" 


6676 FOR I = 1 TO 4: PRINT #2, : NEXT 
6678 PRINT #2, II MULTIPLE-CHANNEL REGION" 
6680 PRINT #2, 


B-32 


6682 PRINT #2, USING " SLOPE = #.#######"; SLO 
PE 
6684 PRINT '2, USING " N-VALUE = ,.#######"; NVA 
LUE 
6685 IF QHMULT(l) < 1 THEN GOTO 6744 

6686 PRINT '2, 

" 

" 

6688 PRINT 12, : PRINT t2, 

6690 PRINT '2, " PROBABILITY OF DISC 
HARGE" 
6692 PRINT f2, " BEING EXCEEDED AT 
THEil 
6694 PRINT f2, " ENERGY DEPTH DISCHARGE APEX BY: 

WIDTH" 
6696 PRINT '2, USING" (FT) (FT) (CFS) 

,.#### (FT)"; BBBB 
6698 PRINT #2, USING" Q t##. 
ff## Q"; 10 A AAAA 
6700 PRINT #2, 
6702 NN -NHMULT 

6704 FORI =1TO NN 

6706 DEPTH = .09168 • NVALUE .6 • QHMULT(I) A .36 / SLOPE A .3

A 

6708 PRINT #2, USING" t#., t#.t ####### '.tttt# 
,.##### #########"; HMULT(I); DEPTH; QHMULT(I); PHYMULT(I); PHZMULT(I 
); WHMULT (I) 
6710 NEXT 
6712 FOR I = 1 TO 4: PRINT #2, : NEXT 
6714 PRINT '2, " 


.. 

6716 PRINT t2, : PRINT #2, 

6717 IF NVMULT = 0 THEN PRINT #2, USING" VELOCITIES BETW 

EEN ##.f AND ##.# FT/SEC"; VMIN; VMAX 
6718 IF NVMULT = o THEN GOTO 6740 
6719 PRINT #2, " PROBABILITY OF DISC 
HARGE" 
6720 PRINT #2, " BEING EXCEEDED AT 
THE" 
6722 PRINT #2, "VELOCITY DEPTH DISCHARGE APEX BY: 

WIDTH" 
6724 PRINT #2, USING " (FT/SEC) (FT) (CFS) 

,.'### (FT)"; BBBB 
6726 PRINT #2, USING" Q ###. 
#### Q"; 10 A AAAA 
6728 PRINT #2, 

6730 NN = NVMULT 

6732FORI-1TONN 
6734 DMULT = .09168 • NVALUE A .6 • QVMULT(I) A .36 / SLOPE A .3 
6736 PRINT #2, USING" ##.# ##.# ####### f.##### 
#.##t#t #########"; VMULT(I); DMULT; QVMULT(I); PVYMULT(I); PVZMULT(I 
); WVMULT (I) 
6738 NEXT 
6740 PRINT #2, CHR$(12) 
6742 RETURN 
6744 PRINT #2, 
6745 PRINT #2," DEPTHS GREATER THAN 0.5 FOOT HAVE PROBABILITIES LESS 

THAN .01" 
6746 RETURN 

8-33 


8200 REM 
8202 REM ************************************************************ 


8204 REM DRAW FLOOD FREQUENCY CURVE 

8206 REM **********************************************************~~ 


8208 REM 

8209 CLS 

8212 LOCATE 1 

8214 PRINT " FLOOD FREQUENCY CURVE" 

8215 NL$ = "NL10;" 

8216 NR$ = "NR10;" 

8217 BOT$ = "R25;NU5;R31;NU5;R42;NU5;R52;NU5;R52;NU5;R42;NU5;R31;NU5;R25; 

" 

8218 TOP$ = "L25;ND5;L31;ND5;L42;ND5;L52;ND5;L52;N05;L42;N05;L31;N05;L25; 

" 

8220 MAG = INT(MU + KG(28) * SIGMA) 

8222 PSET (170, 160) . 

8224 DRAW "X" + VARPTR$(BOT$) 

8226 FORJ = 1 TO MAG 

8228 FORI =2TO10 

8230 UP = INT(144 * LOG(I / (I -1» / LOG(10) / MAG + .5) 

8232 DRAW "U=" + VARPTR$(UP): DRAW "X" + VARPTR$(NL$) 

8234 NEXT I 

8236 NEXT J 

8238 DRAW "X" + VARPTR$(TOP$) 

8240 FOR J := 1 TO MAG 

8242 FORI =2TO10 

8244 OWN = INT(144 * LOG«12 -I) / (11 -I» / LOG(10) / MAG +. 

5) 

8246 DRAW "D=" + VARPTR$ (OWN): DRAW "X" + VARPTR$ (NR$) 

8248 NEXT I 

8250 NEXT J 

8252 LOCATE 22 

8254 PRINT " .999 .99 .9 .5 .1 .01 .001" 

8256FORI =1TO30 

8258 XX = 170 + INT(300 * (KO(30) + KO(I» / KO(30) / 2) 

8260 YY = 160 -INT(144 * (MU + KG(I) * SIGMA -1) / MAG) 

8262 IF YY >160 OR YY < 16 THEN GOTO 8266 

8264 PSET (XX, YY) 

8266 NEXT I 

8268 LOCATE 20, 18 

8270 PRINT 10 

8272 LOCATE 3, 18 -MAG 

8274 PRINT 10 • (MAG + 1) 

8276 LOCATE 5, 28: PRINT "MEAN ="; MU 

8278 LOCATE 6, 25: PRINT "STD DEV ="; SIGMA 

8280 LOCATE 7, 28: PRINT "SKEW ="; SKEW 

8282 IF PDFOPT = 1 THEN GOTO 8286 

8284 LOCATE 8, 24: PRINT "COR COEF ="; RMAX 

8286 RETURN 

8-34 



.8400 REM 
8402 REM ************************************************************ 

8404 REM DRAW FAN 
8406 REM ************************************************************ 


8408 REM 
8410 LOCATE 25 
8412 INPUT H ******** PRESS ENTER TO CONTINUE ******* 
*", KFM 


8414 CLS 

8418 FOR I = 1 TO 4: PRINT: NEXT I 

8420 PRINT B$ 
8422 PRINT 
8424 IF MULT = 6 THEN PRINT "MULTIPLE-CHANNEL REGION" ELSE PRINT "SINGLE


CHANNEL REGION" 

8426 PI = 3.141593 

8427 IF MULT = 6 THEN NH = NHMULT ELSE NH = NHSING 

8428 FOR K=1 TO NH 

8430 IF MULT = 6 THEN W = WHMULT(K) / WHMULT(l) ELSE W = WHSING(K) / W 
HSING(l) 
8432 R =W * 600 

8433 X = INT(R * COS(-PI / 10) / 8): Y = INT((R * SIN(PI / 24) + 102) 

/ 8) 

8434 CIRCLE (0, 90), R, 1, -19 * PI/ 10,-PI/ 10 

8435 LOCATE Y, X: PRINT K -.5 
8436 NEXT K 
8437 IF MULT = 6 THEN NV = NVMULT ELSE NV = NVSING 


8438 FOR K=1 TO NV 

8439 IF MULT = 6 THEN W = WVMULT(K) / WHMULT(l) ELSE W = WVSING(K) / W 
HSING(l) 

8440 R = W * 600 

8441 X = INT(R * COS(-PI/ 10) / 8) -1 

8442 IF X< 0 OR X=0 THEN X=1 

8443 Y = INT((-R * SIN(PI / 24) + 98) /8):IFY < 1 THEN Y =1 

8444 A =19 * PI/ 10 

8446 B = A + 3 * PI/ R 

8448 IF A > 2 * PI THEN A = A-2 * PI 

8450 IF B > 2 * PI THEN B = B-2 * PI 

8452 IFA>PI/ 10ANDA<PITHENGOTO8461 
8454 IF B > PI/ 10ANDB < PI THEN B = PI/ 10 
8456 CIRCLE (0, 90), R, 1, A, B 
8458 A =B + 4 * PI/ R 


8460 GOTO 8446 
8461 LOCATE Y, X 
8462 IF MOLT = 6 THEN PRINT K + VMULT(l) -1 ELSE PRINT K + 2.5 
8463 NEXT K 
8464 LOCATE 18 
8466 PRINT " DEPTH" 
8468 PRINT 
8470 PRINT "----VELOCITY" 
8472 MOLT = MULT + 1 
8474 IF MOLT = 6 AND QHMULT(l) > 1 THEN GOTO 8410 
8476 LOCATE 25 
8478 INPUT II ******** PRESS ENTER TO CONTINUE ******* 

*", EMM 
8480 RETURN 


8-35 



9000 REM ************************************************************ 

9002 REM OPTION TO VIEW ANDIOR PRINT OUTPUT DATA 
9006 REM ********************************************************* 


9008 A$ = n II + A$ 
9010 SCREEN 0: COLOR 12 
9020 CLS 

9022 FOR I = 1 TO 8: PRINT: NEXT I 

9024 INPUT II DO YOU WISH TO VIEW SOME OUTPUT DATA (YIN)? 

", V$ 
9026 CLS : COLOR 10 
9027 IF INSTR(V$, "Y") = 0 AND INSTR(V$, lIyll) = 0 THEN GOTO 9900 

9028 FOR I =1 TO 5: PRINT : NEXT 

9030 PRINT 
9032 PRINT PRINT : COLOR 10 
9034 PRINT II (l) •••• FLOOD FREQUENCY DATA" 
9036 PRINT 
9038 PRINT" (2) •••• TRANSFORMATION DATA" 
9040 PRINT 
9042 PRINT (3) •••• 100-YEAR DEPTH-ZONE DATA SINGLE-CHANN

II 

EL REGION" 
9044 PRINT 
9046 PRINT " (4) •••• 100-YEAR VELOCITY-ZONE DATA --SINGLE-CH 
ANNEL REGION" 
9048 IF MULT < 5.5 THEN GOTO 9056 ELSE PRINT 
9050 PRINT" (5) .••• 100-YEAR DEPTH-ZONE DATA --MULTIPLE-CHA 
NNEL REGION" 
9052 PRINT 
9054 PRINT " (6) •••• 100-YEAR VELOCITY-ZONE DATA --MULTIP~ 
-CHANNEL 
REGION" 
9056 PRINT CHR$(ll): FOR I = 1 TO 2: PRINT CHR$(31): NEXT I 
9057 COLOR 12 
9058 INPUT II PLEASE SELECT, BY NUMBER, THE DATA THAT YOU WISH TO VI 
EW••••·, SEL 
9059 COLOR 10 
9060 IF SEL = 1 THEN GOTO 9100 
9062 IF SEL = 2 THEN GOTO 9200 
9064 IF SEL = 3 THEN GOTO 9300 
9066 IF SEL = 4 THEN GOTO 9400 
9067 IF MULT < 5.5 THEN GOTO 9072 

9068 IF SEL = 5 THEN GOTO 9500 

9070 IF SEL = 6 THEN GOTO 9600 
9072 CLS : PRINT : PRINT 
9074 PRINT SORRY, THERE IS NO DATA SELECTION NUMBER "; S

II 

EL 

9076 FOR I =1 TO 5: PRINT : NEXT I 

9077 COLOR 12 
9078 GOTO 9024 

B-36 



9100 REM 
9102 REM ************************************************************ 


9104 REM FLOOD FREQUENCY OUTPUT 
9106 REM ************************************************************ 


9107 REM 
9108 NO = MOY: SIGMA = SIGMAY 
9109 CLS : PRINT 
9110 PRINT A$ 
9116 FOR I = 1 TO 2: PRINT: NEXT 
9118 IF PDFOPT = 2 GOTO 9124 
9120 PRINT" FLOOD FREQUENCY CURVE DEFINED BY MEAN, STANDARD DEVIA 
TION, AND SKEW" 
9122 GOTO 9140 
9124 PRINT " FLOOD FREQUENCY CURVE DEFINED BY LEAST-SQUARES FIT OF 

DATA" 

9126 FOR I = 1 TO 2: PRINT: NEXT 

9128 PRINT " RETURN INTERVAL INPUT DISCHARGE BEST FIT DISCHA 
RGE" 
9130 PRINT " (YEARS) (CFS) (CFS) " 
9132 PRINT 
9134 FOR K = 1 TO NOF 
9136 PRINT USING " #### ###### 
"###f": RET(K): Q(K): 10 (SIGMA * KK(K) + MO)


A 

9138 NEXT 
9139 GOTO 9180 
9140 PRINT 
9142 PRINT USING " MEAN = ##. ######" 
: HOY 
9144 PRINT USING" STANDARD DEVIATION = ##.######" 
; SIGMAY 
9146 PRINT USING " SKEW = ##. #": SKE 
W 
9148 FOR I = 1 TO 2: PRINT: NEXT 

9150 PRINT" SUMMARY OF DISCHARGES:" 
9152 PRINT 
9154 PRINT USING" 10-YEAR DISCHARGE = ######": 1 

o A (SIGMA * KG(20) + MO) 
9156 PRINT USING" 50-YEAR DISCHARGE = ######": 1 

o A (SIGMA * KG(24) + MO) 
9158 PRINT USING" 100-YEAR DISCHARGE = ######": 1 
o A (SIGMA * KG(25) + MO) 
9160 PRINT USING" 500-YEAR DISCHARGE = ######"; 1 

o A (SIGMA * KG(27) + MO) 
9161 IF INSTR(FFD$, "Y") = 1 OR INSTR(FFD$, "y") = 1 THEN GOTO 9192 
9162 FOR I = 1 TO 3: PRINT: NEXT I 
9163 COLOR 12 
9164 INPUT" DO YOU WISH TO VIEW MORE OUTPUT DATA (YIN)? ", V$ 
9168 GOTO 9026 
9180 FOR I = 1 TO 3: PRINT: NEXT I 
9181 COLOR 12 
9182 INPUT " ******** PRESS ENTER TO CONTINUE ******* 
*", EMM 
9183 CLS : COLOR 10 

9184 FOR I =1 TO 4: PRINT NEXT I 

9185 PRINT A$ 

9186 FOR I =1 TO 3: PRINT NEXT I 

B-37 


9188 PRINT n FLOOD FREQUENCY CURVE DEFINED BY LEAST-SQUARES FIT OF DAT 
A" 

9190 GOTO 9140 
9192FORI..1TO5:PRINT: NEXTI 
9194 PRINT " STANDARD DEVIATION TIMES SKEW GREATER THAF 
2.1" 

9195 PRINT " WIDTHS CANNOT BE COMPUTED" 
9196 PRINT : COLOR 12 
9197 PRINT " PROGRAM TERMINATED" 
9198 PRINT : INPUT " PRESS ENTER TO CONTINU 
E", RTP 
9199 GOTO 9900 

9200 REM ************************************************************ 

9201 REM TRANSFORMATION OUTPUT 
9202 REM ************************************************************ 


9203 CLS 

9204 FOR I = 1 TO 4: PRINT: NEXT 

9205 PRINT A$ 
9206 REM 
9208 FOR I = 1 TO 2: PRINT: NEXT 
9210 IF SKEW = 0 GOTO 9218 

9212 AAAA = -.92 * TRANS I (SCALE -.92): BBBB = SCALE I (SCALE -.92)

9214 PRINT USING" STATISTICS AFTER TRANSFORMATION OF Y=LOG(Q) TO Z= 
#.####+f.#### LOG(Q)": -.92 * TRANS I (SCALE -.92): SCALE I (SCALE -.92 

) 

9216 GOTO 9222 
9218 AAAA = .92 * SIGMAY * SIGMAY: BBBB = 11 

9220 PRINT USING" STATISTICS AFTER TRANSFORMATION OF Y=LOG(Q) TO 
#.#f##+LOG(Q)"; .92 * SIGMAY * SIGMAY 
9222 PRINT 
9224 PRINT USING" MEAN OF Z -##.#ttt#t": MU 

Z 

9226 PRINT USING" STANDARD DEVIATION = tt.tt#ttt": SI 
GMAZ 
9228 PRINT USING" SKEW" t#. #####t": SK 
EW 
9230 PRINT USING" TRANSFORMATION CONSTANT" ##.######": CN 
ST 
9232FORI=1TO5:PRINT NEXTI 
9233 COLOR 12 
9236 INPUT" DO YOU WISH TO VIEW MORE OUTPUT DATA (YIN)? ", V$ 
9238 GOTO 9026 

8-38 



9300 REM 
9301 REM ************************************************************ 


9302 REM DEPTH-ZONE OUTPUT DATA 
9304 REM ************************************************************ 


9306 CLS 

9308 FOR I 1 TO 2: PRINT NEXT IK 

9310 PRINT A$ 
9312 FOR I = 1 TO 2: PRINT NEXT 
9314 PRINT .. SINGLE-CHANNEL REGION" 
9316 PRINT 


It 

" 

9318 PRINT PRINT 
9320 PRINT PROBABILITY OF DISCH

It 

ARGE" 

9322 PRINT .. BEING EXCEEDED AT 
THEil 
9324 PRINT ENERGY DEPTH DISCHARGE APEX BY:

II 

WIDTH" 
9326 PRINT USING (FT) (FT) (CFS)

It 

,.#### (FT)"; BBBB 
9328 PRINT USING " Q

",# Q"; 

10 AAAA

A 

9330 PRINT 
9332 NN -NHSING 
9334 FOR I = 1 TO NN 

9336 DEPTH 2 * HSING(I) I 3

K 

9338 PRINT USING" ##.# ##.# ####1## #.#,### 
,. ###1# #########"; HSING (I); DEPTH; QHSING (I) ; ·PHYSING (I); PHZSING (I); waSING(I) 
9340 NEXT 
9341 PRINT 

.It.

II

9342 PRINT AVULSION FACTOR , AVUL 

9343 FOR I = 1 TO 3: PRINT: NEXT I 

9344 COLOR 12 
9346 INPUT DO YOU WISH TO VIEW MORE OUTPUT DATA (YIN)? ", V$


It 

9348 GOTO 9026 

8-39 



9400 REM 
9401 REM ************************************************************ 


9402 REM VELOCITY-ZONE OUTPUT DATA 
9404 REM *********************************************************. 


9406 CLS 

9408 FOR I -1 TO 2: PRINT: NEXT I 

9410 PRINT A$ 

9412 FOR I = 1 TO 2: PRINT: NEXT I 

9420 PRINT SINGLE-CHANNEL REGION"

II 

9422 PRINT " 

" 

9424 PRINT PRINT 
9426 PRINT " PROBABILITY OF DISC 
HARGE" 
9428 PRINT " BEING EXCEEDED AT 
THE" 
9430 PRINT " VELOCITY DEPTH DISCHARGE APEX BY: 


WIDTH" 
9432 PRINT USING " (FT/SEC) (FT) (CFS)

f.tttf (FT)"; BBBB 
9434 PRINT USING " Q fff. 
ffff Q"; 10 A AAAA 
9436 PRINT 
9438 NN -NVSING 
9440 FOR I = 1 TO NN 
9442 DSING -VSING(I) A 2 I 32.16 
9444 PRINT USING " ff.f ff.f ffttttf f.fffff 

f.fftff fffffffff"; VSING(I); DSING7 QVSING(I); PVYSING(I); PVZSINr'~ 
); WVSING(I) 
9446 NEXT 
9447 PRINT 
9448 PRINT " AVULSION FACTOR -"; AVUL 

9449 FOR I =1 TO 3: PRINT : NEXT I 

9450 COLOR 12 
9452 INPUT " DO YOU WISH TO VIEW MORE OUTPUT DATA (YIN)? ", V$ 
9454 GOTO 9026 

8-40 



9500 REM 
9501 REM ************************************************************ 


,-9502 REM DEPTH-ZONE OUTPUT DATA 
9504 REM ************************************************************ 

9506 CLS 

9508 FOR I =1 TO 2: PRINT NEXT I 

9510 PRINT A$ 
9512 FOR I = 1 TO 2: PRINT NEXT I 
9520 PRINT " MULTIPLE-CHANNEL REGION" 
9530 PRINT 
9532 PRINT USING" SLOPE -f.fffffff"; SLO 
PE 
9534 PRINT USING" N-VALUE -f.ffffiff"; NVA 

WE 
9536 IF QHMULT(l) < 1 THEN GOTO 9574 
9538 PRINT " 


" 

9540 PRINT : PRINT 
9542 PRINT " PROBABILITY OF DISC 
HARGE" 
9544 PRINT " BEING EXCEEDED AT 
THE" 

9546 PRINT " ENERGY DEPTH DISCHARGE APEX BY: 
WIDTH" 
9548 PRINT USING " (FT) (FT) (CFS)

f.ft'f (FT)"; BBBB 
9550 PRINT USING " Q 
flil Q"; 10 AAAA

A 

9552 PRINT 
9554 NN -NHMULT 
9556FORI-1TONN 

9558 DEPTH = .09168 * NVALUE A .6 * QHMULT(I) A .36 I SLOPE A .3 

9560 PRINT USING " ft.t ft.f ttttttt t.fttff 
I."'f' ffffffff#"; HMULT (I); DEPTH; QHMULT (I); PHYMULT (I); PHZMULT (I 
); WHMULT(I) 
9562 NEXT 
9564 GOTO 9577 
9574 PRINT 
9576 PRINT n DEPTHS GREATER THAN 0.5 FOOT HAVE PROBABILITIES LESS 

THAN .01" 
9577 PRINT 
9578 PRINT n AVULSION FACTOR -"; AVUL 
9580 FOR I -1 TO 3: PRINT: NEXT I 
9584 COLOR 12 
9586 INPUT " DO YOU WISH TO VIEW MORE OUTPUT DATA (YIN)? It, V$ 
9588 GOTO 9026 

8-41 



9600 REM 
9601 REM ********************************************************** . 


9602 REM VELOCITY-ZONE OUTPUT DATA 
9604 REM ************************************************************ 


9606 ·CLS 

" '.,.: 

9608 FOR lei TO 2: PRINT: NEXT I 
9610 PRINT A$ 
9612 FOR I -= 1 TO 2: PRINT NEXT I 
9620 PRINT " MULTIPLE-CHANNEL REGION·II. 
9622 PRINT : PRINT 
9624 IF NVMULT = 0 THEN PRINT USING tI VELOCITIES BETW 
EEN tt.t AND tt.t FT/SEC": WIN: VMAX 
9626 IF NVMULT = 0 THEN GOTO 9650 
9628 PRINT " PROBABILITY OF DISC 
HARGE" 
9630 PRINT n BEING EXCEEDED AT 
THE" 
9632 PRINT n VELOCITY DEPTH DISCHARGE APEX BY: 

WIDTH" 
9634 PRINT USING tI (FT/SEC) (FT) (CFS)
'.'ttl (FT)": BBBB 


9636 PRINT USING Q

tI 

til' Q"; 10 A AAAA 
9638 PRINT 
9640 NN -NVMULT 
9642 FOR I = 1 TO NN 


9644 DMULT = .09168 * NVALUE A .6 * QVMULT(I) A .36 I SLOPE A .3 
9646 PRINT USING tI tl.1 tt.t tttlitt t.ttttl 
, ••,tll tttlilltl": VMULT (I): DMULT; QVMULT (I); PVYMULT (I); PVZMULl \ .i 
); WVMULT(I) 
9648 NEXT 
9649 PRINT 

.tI.

tI

9650 PRINT AVULSION FACTOR , AVUL 

9651 FOR I = 1 TO 3: PRINT: NEXT I 
9652 COLOR 12 
9654 INPUT " DO YOU WISH TO VIEW MORE OUTPUT DATA (YIN)? .. , V$ 
9656 GOTO 9026 


9900 REM 

************************************************************* 

***** 

9902 REM OPTION TO PRINT OUTPUT 
9904 REM 


************************************************************* 

***** 

9906 REM 
9908 REM 
9910 COLOR 12 
9912 FOR I -= 1 TO 4: PRINT: NEXT I 


II ,

9914 INPUT" DO YOU WISH TO PRINT THE OUTPUT (YIN)?

PRNT$ 
9916 IF INSTR(PRNT$, tly") = 0 AND INSTR(PRNT$, lIyll) • 0 THEN GOTO 9920 
9918 OPEN "FANN" FOR OUTPUT AS #3 
9920 CLS 
9923 SYSTEM 

8-42 



1 REM 
2 REM AAA GGGGGGGGG AAA 111111111 
3 REM AAAAA GGG GGG AAAAA III 

.,',: ., 
4 REM AAA AAA GGG AAA AAA III 
5 REM AAA AAA GGG AAA AAA III 
6 REM AAAAAAAAAAAA GGG GGGGGG AAAAAAAAAAAA III 
7 REM AAA AAA GGG GGG AAA AAA· III 
8 REM AAA AAA GGGGGGGGG AAA AAA 111111111 
9 REM 

10 CLS 
20 COLOR 12 
30FOR1=1TO6 :PRINT:NEXTI 
40 INPUT n DO YOU WISH TO MAKE ANOTHER RUN 
N$ 
50 IF INSTR(NWRN$,"yll)=O AND INSTR(NWRN$,"y")=O THEN GOTO 70 
60 OPEN "FANN" FOR OUTPUT AS #3 
70 SYSTEM 

; I. 

NNN NNN 
NNNN NNN 
NNNNNN NNN 
NNN NNN NNN 
NNN NNNNN 
NNN NNNN 
NNN NNN 

.,' ". 

(YIN)? n ,NWR 

"1'-" 

" 'r... 

8-43