HP 32SII: Tracing the Orbit Coordinates

HP 32SII:  Tracing the Orbit Coordinates

The following program tracks the position (x,y) of a satellite in orbit by solving the pair of second-order differential equations:

d^2 x/dt^2 = -α * x * (x^2 + y^2)^-1.5

d^2 y/dt^2 = -α * y * (x^2 + y^2)^-1.5

For planetary orbits, α = G * M

Variables and Labels

Labels Used:  4 (A, B, C, D)

Variables Used:

Pre stored variables:

X = x0, initial x position

U = dx/dt, velocity of x coordinate

Y = y0, initial y position

V = dy/dt, velocity of y coordinate

C = Δt, desired time increment

A = α, gravitational force factor

If you are analyzing planetary orbits, then α = G * M where M is the mass of the center object and G is the Universal gravitational constant, G = 6.67384 * 10^-11 (m^3)/(s^2 * kg). 

Other:

T = t

This program is based off of the HP 25S and program authored Robert M. Eisberg. (see the source below)

HP 32SII Program (87.5 bytes)

A01 LBL A   // start of program

A02 0

A03 STO T   // initialize time variable

C01 LBL C

C02 RCL Y

C03 RCL Y

C04 x^2

C05 RCL X

C06 x^2

C07 +

C08 -1.5

C09 y^x

C10 RCL A

C11 *

C12 *       // calculate –α*y*(x^2 + y^2)^-1.5

C13 LASTx 

C14 RCL X

C15 *      // calculate –α*x*(x^2 + y^2)^-1.5

C16 RCL T

C17 x=0?

C18 GTO B

C19 R↓

D01 LBL D

D02 RCL C

D03 *

D04 STO- U

D05 R↓

D06 RCL C

D07 *

D08 STO- V

D09 RCL C

D10 STO+ T

D11 RCL U

D12 *

D13 STO+ X

D14 RCL X

D15 STOP       // view X

D16 RCL C

D17 RCL V

D18 *

D19 STO+ Y

D20 RCL Y

D21 STOP       // view Y

D22 GTO C      // start a new loop

B01 LBL B

B02 R↓

B03 2

B04 ÷

B05 x<>y

B06 2

B07 ÷

B08 x<>y

B09 GTO D

During execution, recall T for the time variable.  Run by XEQ A.

Example

Pre stored variables:

X = 1.5

U = 0.18

Y = 1.5

V = 0.23

C = 1

A = α = 1

Table:

T =	X =	Y =
0   (initial)	1.5	1.5
1	1.6014	1.6514
2	1.5713	1.6672
3	1.4105	1.5443
4	1.0956	1.2527
5	0.5429	0.6892
6	-0.8138	-0.8949

Source:  Robert M. Eiseberg.  Applied Mathematical Physics with Programmable Calculators McGraw Hill.  1976.  ISBN 0-07-019109-3

Eddie

This blog is property of Edward Shore, 2017.

Geometry: The Intersection Point of a Quadrilateral

Geometry: The Intersection Point of a Quadrilateral

The Setup

We are given four points, A, B, C, and D, as four Cartesian coordinates.  We connect the four points, starting with A, in a clockwise form to form a quadrilateral.  We will designate each points as the coordinates:

A:  (ax, ay)

B:  (bx, by)

C:  (cx, cy)

D:  (dx, dy)

Draw a line from one corner to the opposite corner.  This results in two lines: AC and DB.  The two lines (show above in lime green) meet at point P.  The goal is determine the coordinates of P.

Slope

We know the equation of the line is y = m*x + b, where m is the slope and b is the y-intercept. Note that:

y = m*x + b

y – m*x = b

In geometry, the slope of a line containing two points is generally defined as:

m = (change in y coordinates)/(change in x coordinates) = Δy/Δx = (y2 – y1)/(x2 – x1)

The slope of AC:   SAC = (cy – ay)/(cx – ax)

The slope of BD:  SBD = (dy – by)/(dx –bx)

The Intercept

We can easily deduce that solving the general equation of the line y = m*x + b for the intercept yields b = y – m*x.  If follows that:

The intercept of AC:  IAC = cy – SAC * cx = ay – SAC *ax

The intercept of BD:  IBD = by – SBD * bx = dy – IBD *dx

Finding the Intersection Point

Now that the slopes and intercepts are determined, we can form the following system of equations:

(I) y = SAC * x + IAC

(II) y = SBD * x + IBD

Solving for x and y will find our intersection point P (px, py).  Subtracting (II) from (I) (see above):

0 = (SAC – SBD) * x + (IAC – IBD)

We can solve for x.

-(IAC – IBD) = (SAC – SBD) * x

(-1*IAC - -1*IBD) = (SAC – SBD) * x

(-IAC + IBD) = (SAC – SBD) * x

(IBD – IAC) = (SAC – SBD) * x

Hence:

x = px = (IBD – IAC)/(SAC – SBD)

It follows that we determine y by either equation (I) or (II):

y = py = SAC * px + IAC = SBD * px + IBD

Our desired coordinates of point P are found.

Eddie

This blog is property of Edward Shore, 2017.

x

Casio fx-3650p and HP 21S: Minimum Loss Matching

Casio fx-3650p and HP 21S: Minimum Loss Matching

This program takes the incoming impedances Z0 (from left) and Z1 (from right), and determines the resistance R1 and R2, with the corresponding minimum loss.  The program assumes that Z0 and Z1 are both positive and Z0 ≥ Z1. 

Formulas Used:

R1 = Z0 * √(1 – Z1/Z0)

R2 = Z1 / √(1 – Z1/Z0)

Minimum loss = 20 * log (√(Z1/Z0) + √(Z1/Z0 -1))

Casio fx-3650P

Input:  X = Z0, Y = Z1

Program (49 steps):

? → X : ? → Y : Y ÷ X → M :

X √ ( 1 – M → A ◢

Y √ ( 1 – M → B ◢

20 log ( √ M ^-1 + √ ( M ^-1 – 1 → C

HP 21S

Input:  Z0 gets stored in register 0, Z1 gets stored in register 1

Program (44 steps):

Step	Code	Key	Notes
01	61, 41, C	LBL C	Start the program
02	22, 1	RCL 1	Z1
03	45	÷
04	22, 0	RCL 0	Z0
05	74	=
06	21, 4	STO 4
07	33	(
08	1	1
09	65	-
10	22, 4	RCL 4
11	34	)
12	11	√
13	21, 5	STO 5
14	55	*
15	22, 0	RCL 0
16	74	=
17	21, 2	STO 2
18	26	R/S	Display R1
19	22, 1	RCL 1
20	45	÷
21	22, 5	RCL 5
22	74	=
23	21, 3	RCL 3
24	26	R/S	Display R2
25	33	(
26	22, 4	RCL 4
27	15	1/x
28	11	√
29	75	+
30	33	(
31	22, 4	RCL 4
32	15	1/x
33	65	-
34	1	1
35	34	)
36	11	√
37	34	)
38	51, 13	LOG
39	55	*
40	2	2
41	0	0
42	74	=
43	21, 5	STO 5	Display minimum loss (Z)
44	61, 26	RTN

Example:

Input:  Z0 = 50, Z1 = 45

Results: R1 = 15.8113883008, R2 = 142.302494708, Min Loss = 2.84419586692

Source:  Casio Scientific Formula 128 fx-1000F/fx-5000F owner’s manual, 1987

Eddie

This blog is property of Edward Shore, 2017