HP 15C: Static
Equivalent at a Point
Source: Step by Step
Solutions for your HP Calculator: Engineering Applications. HP32S.
Hewlett Packard. Corvallis,
OR. 1988
pg. 4044
Equations:
R5 * cos θ1 + R6 * cos θ2 =  Σ F * cos ϕ
R5 * sin θ1 + R6 * sin θ2 =  Σ F * sin ϕ
This program solves for R5, R6.
Variables:
Inputs:
R0: number of points
R1: T = angle of each
known force (ϕ)
R2: F = value of each
known force (F)
A = direction of the first reaction force (θ1)
B = direction of the second reaction force (θ2)
Outputs:
R5: R1 = unknown force 1
(R1)
R6: R2 = unknown force 2
(R2)
Temporary Variables:
R3, R4, R7, R8, R9, R.0
R3 = X = sin θ1
R9 = Y = cos θ1
R4 = C = sin θ2
R.0 = D = cos θ2
R7 = Σ F * cos ϕ
R8 = Σ F * sin ϕ
This program clears all the registers.
Instructions:
Enter n, press [ f ]
[1/x] ( E ), then each pair of T then [R/S] and F then [R/S],
respectively. When completed with the
pair, enter A and B. R1 is displayed
first. Press [R/S] to get R2.
Caution:
The HP 15C does not have an alpha numeric display, so you
have to keep track of everything yourself.
For register .0, press [ . ] [ 0 ].
Program:
Step

Key

Key Code

001

LBL E

42, 21, 15

002

Clear Registers [ f ] [X<>Y]

42, 34

003

STO 0

44, 0

004

LBL 1

42, 21, 1

005

R/S (enter T here)

31

006

STO 1

44, 1

007

R/S (enter F
here)

31

008

STO 2

44, 2

009

>R (Polar to
Rectangular)

42, 1

010

STO+ 7

44, 40, 7

011

X<>Y

34

012

STO+ 8

44, 40, 8

013

DSE 0

42, 5, 0

014

GTO 1

22, 1

015

R/S (enter A here)

31

016

SIN

23

017

STO 3

44, 3

018

LST X

44, 36

019

COS

24

020

STO 9

44, 9

021

R/S (enter B here)

31

022

SIN

23

023

STO 4

44, 4

024

LST X

43, 36

025

COS

24

026

STO .0 (note
the .0)

44, .0

027

RCL 7

45, 7

028

RCL* 4

45, 20, 4

029

RLC .0

45, .0

030

RCL* 8

45, 20, 8

031



30

032

RCL 3

45, 3

033

RCL* .0

45, 20, .0

034

RCL 9

45, 9

035

RCL* 4

45, 20, 4

036



30

037

÷

10

038

STO 5

44, 5

039

R/S (display R1)

31

040

LST X

43, 36

041

RCL 9

45, 9

042

RCL* 8

45, 20, 8

043

RCL 7

45, 7

044

RCL* 3

45, 20, 3

045



30

046

X<>Y

34

047

÷

10

048

STO 6

44, 6

049

RTN

43, 32

Examples:
Balance a vector with length 85 turned at 144°. The reactionary direction forces are 45° and
270°. In this case, n = 1, F = 85, T =
144, A = 45, and B = 270.
Key strokes:
1 [ f ] [ 1/x ] (LBL
E)
144 [ R/S ] (T)
85 [ R/S ] (F)
45 [ R/S ] (A)
270 [ R/S ] (B)
Results:
R5 = 97.2504, press
[R/S]
R6 = 118.7282
Truss:
n = 2:
T1 = 0°, F1 = 176
T2 = 135°, F2 = 100
A = 45°, B = 180°
Key strokes:
2 [ f ] [ 1/x ] (LBL
E)
0 [ R/S ] (T1)
176 [ R/S ] (F1)
135 [ R/S ] (T2)
100 [ R/S ] (F2)
45 [ R/S ] (A)
180 [ R/S ] (B)
Results:
R5 = 100 (R5), press [ R/S ]
R6 = 37.5786
I wonder would be preferable, one blog entry for each subject or one blog post that captures multiple subjects for one calculator. Comments are welcome.
Eddie
This blog is property of Edward Shore. 2016