TI 30Xa Algorithm: Acceleration, Velocity, Speed

TI 30Xa Algorithm: Acceleration, Velocity, Speed

Introduction and Algorithm

Given the acceleration (α), initial velocity (v0), and initial position (p0) of an object on a 2D plane, calculate velocity and position at time T.

Assuming that acceleration is constant, the equations are:

Acceleration: A = α

Velocity: V = α * t + v0

Position: P = α / 2 * t^2 + v0 * t + p0

With some simplification, we can simplify the work flow:

( I )

A = α

( II )

V = α * t + v0

V = A * t + v0 (II)

( III )

P = α / 2 * t^2 + v0 * t + p0

P = t * ( α / 2 * t + v0 ) + p0

P = t * ( α / 2 * t + 2 * v0 / 2 ) + p0

P = t * ( (α * t + 2 * v0) / 2 ) + p0

P = t / 2 * ( α * t + 2 * v0 ) + p0

P = t / 2 * ( α * t + v0 + v0 ) + p0

P = t * (V + v0) / 2 + p0

( I ): A = α

( II ): V = A * t + v0

( III ): P = t * (V + v0) / 2 + p0

Key Strokes:

1. Store the time T in Memory 1 ( T [ STO ] 1 )

2. Store the initial velocity in Memory 2 ( v0 [ STO ] 2 )

3. Enter the acceleration constant, α, press [ = ]

4. Calculate the velocity at time T: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ]

5. Calculate the position at time T: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] p0 [ = ]

Examples

Example 1:

α = 1.3

v0 = 0

p0 = 0

T = 10

1. 10 [ STO ] 1

2. 0 [ STO ] 2

3. Acceleration: 1.3 [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: 13)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 0 [ = ] (Result: 65)

Example 2:

α = -9.80665

v0 = 0

p0 = 10000

T = 30

1. 30 [ STO ] 1

2. 0 [ STO ] 2

3. Acceleration: 9.80665 [ +/- ] [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: -294.1995)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 10000 [ = ] (Result: 5587.0075)

Example 3:

α = -0.05

v0 = 10

p0 = 20

T = 15

1. 15 [ STO ] 1

2. 10 [ STO ] 2

3. Acceleration: 0.05 [ +/- ] [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: 9.25)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 20 [ = ] (Result: 164.375)

Example 4:

α = 4.26

v0 = 1.8

p0 = 0

T = 5

1. 5 [ STO ] 1

2. 1.8 [ STO ] 2

3. Acceleration: 4.26 [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: 23.1)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 0 [ = ] (Result: 62.25)

Hope you find this useful.

Happy Halloween, Everyone!

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

HHC 2024 Videos are Up

 HHC 2024 Videos are Up

Hi, everyone.  Just want to let you know that the videos of the talks from the wonderful HHC 2024 conference are now up on hpcalc.org's on YouTube.   

As usual, time flew by and we had a great time.  

https://www.youtube.com/@hpcalc

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Swiss Micros DM32: Reimann-Louiville Fractional Integral of x^p

Swiss Micros DM32: Reimann-Louiville Fractional Integral of x^p

Introduction

The program presented today calculates the Riemann-Louiville integral of:

f(t) = t^p, where p is a real number.

The formula for this integral is:

cDx^(-v) = 1 / Γ(v) * ∫( (x – t) * t^p dt, t = c, t = x)

= ∫( ((x – t) * t^p) / (v -1)! dt, t = c, t = x)

I covered these type of integrals on my September 14, 2024 blog.

DM32 Program: Reimann-Louiville Fractional Integral of x^p

LBL F

INPUT C

INPUT X

INPUT V

INPUT P

FN= I

RCL C

RCL X

∫ FN d T

RTN

LBL I

RCL X

RCL- T

RCL V

1

-

y^x

RCL T

RCL P

y^x

×

RCL V

1

-

x!

÷

RTN

Here is a text version that can be transferred to a dm32 state file (fractionalintegralm.d32):

https://drive.google.com/file/d/1E-wUq4GW5dX06VZ-5WWwy7KRm3SN_uyq/view?usp=sharing

Examples

Run program F: XEQ F. Make sure that V > 0.

C	X	V	P	Result (FIX 5)
0	5	2	2	52.08333
1	6	3	3	385.41667
2	7	1.5	3	717.69103
0	1	1.75	4	0.05298

Source

Kimeu, Joseph M., "Fractional Calculus: Definitions and Applications" (2009).Masters Theses & Specialist Projects. Paper 115. http://digitalcommons.wku.edu/theses/115

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Casio fx-CG 50: Centroid of a 2D spaces

Casio fx-CG 50: Centroid of a 2D spaces

The program CENTROID calculates the center point for an area covered by:

y(x) ≥ 0, x ≥ lower limit, x ≤ upper limit

The center point is calculated by:

x-center = ∫( x * y(x) dx, x = lower limit, x = upper limit) / A

y-center = ∫(1 / 2 * y(x)^2 dx, x = lower limit, x = upper limit) / A

where A = ∫( y(x) dx, x = lower limit, x = upper limit)

Casio fx-CG 50 Program Code: CENTROID

Here is the code, which includes a graphic representation of y(x) and the location of the centroid. In this code, the Y is bold and it comes from the VARS menu. Do not merely use ALPHA+Y. For calculators with monochrome screens, such as the fx-9750G/fx-9860G series, leave out the color commands (Black, Blue, Red). The program assumes that there no preset plots or more than one function to be plotted.

This program works best for functions y(x) ≥ 0 for x ∈ [lower limit, upper limit].

Here is a text-only version:

Examples

Example 1: y = x^2 , lower limit = 0, upper limit = 1

X-Center = 3 / 4 = 0.75

Y-Center = 3 / 10 = 0.3

Example 2: y = 2 * cos( x / 2 ), lower limit = 0, upper limit = π

X-Center ≈ 1.141592654

Y-Center ≈ 0.7853981634

Example 3: y = 3, lower limit = 1, upper limit = 5

X-Center = 3

Y-Center = 1.5

Example 4: y = -4 * x^2 + 2 * x + 6, lower limit = -0.5, upper limit = 1

X-Center = 1 / 4

Y-Center = 307 / 110

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Sum of Sequential Integers (featuring Swiss Micros DM32)

 Sum of Sequential Integers (featuring Swiss Micros DM32)

What is the sum of the following:

100 + 101 + 102 + 103 + 104 + 105 + … + 997 + 998 + 999?

It’s doable on a calculator, but going straight forward will require a lot of keystrokes (unless you have access to the sigma function (Σ)).

Note that:

100 + 101 + 102 + 103 + 104 + 105 + … + 997 + 998 + 999

= 100 + (100 + 1) + (100 + 2) + (100 + 3) + (100 + 4) + …. + (100 + 897) + (100 + 898) + (100 + 899)

= (100 + 0) + (100 + 1) + (100 + 2) + (100 + 3) + (100 + 4) + …. + (100 + 897) + (100 + 898) + (100 + 899)

Notice the sum starts with a base, 100. The sequence goes for 900 terms in the form of 100+n where n = 0 to 899. Let’s look at a general case.

Let x be a base integer, and S be the sum as:

S = (x + 0) + (x + 1) + (x + 2) + (x + 3) + …. + (x + n)

Rearranging the terms leads us to:

S = (x + x + x + x + … + x) + (0 + 1 + 2 + 3 + 4 + … + n)

There are n + 1 pairs:

S = (x * (n + 1)) + (0 + 1 + 2 + 3 + 4 + … + n)

S = (x * (n + 1)) + (1 + 2 + 3 + 4 + … + n)

The sum of Σ( k, k = 1 to k = n) = 1 + 2 + 3 + 4 + … + n = n * (n + 1) / 2

Then:

S = (x * (n + 1)) + n * (n + 1) / 2

Let’s look at some specific examples:

n = 1

S = (x + 0) + (x + 1)

S = (x + x) + (0 +1)

S = 2 * x + 1

n = 2

S = (x + 0) + (x + 1) + (x + 2)

S = (x + x + x) + (0 + 1 + 2)

S = 3 * x + 3

2 * 3 / 2 = 3

n = 3

S = (x + 0) + (x + 1) + (x + 2) + (x + 3)

S = (x + x + x + x) + (0 + 1 + 2 + 3)

S = 4 * x + 6

3 * 4 / 2 = 6

S = 500 + 501 +502 + 503

S = (500 + 0) + (500 + 1) + (500 + 2) + (500 + 3)

S = (500 + 500 + 500 + 500) + (0 + 1 + 2 + 3)

S = ((3 + 1) * 500) + (3 * 4 / 2)

S = 2000 + 6

S = 2006

S = 250 + 251 + 252 + 253 + … + 269 + 270

Base = 250

n = 270 – 250 = 20

Then:

S = 250 * (20 + 1) + 20 * 21 / 2

S = 5460

Let’s go our original problem:

S = 100 + 101 + 102 + 103 + 104 + 105 + … + 997 + 998 + 999

Base = 100

n = 999 -100 = 899

Then:

S = 100 * (899 + 1) + 899 * 900 / 2

S = 90000 + 404550

S = 494550

The program code calculates the sum:

Swiss Micros DM32 Program (also HP 32SII)

S01 LBL S

S02 INPUT X

S03 INPUT N

S04 RCL N

S05 1

S06 +

S07 RCL× N

S08 RCL N

S09 1

S10 RCL+ N

S11 ×

S12 2

S13 ÷

S14 +

S15 RTN

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.