Retro Review: TI-35 Galaxy Solar

Retro Review:   TI-35 Galaxy Solar

Quick Facts

Model: TI-35 Galaxy Solar

Company: Texas Instruments

Production Years:  1984 - 1989 (mainly 1985)

Power:  Solar

Type:  Scientific

Operating System:  AOS

Memory Registers:  1

Display:  One line:  8 digits or 5 digits with 2 digit exponent (10^xx), with polar/rectangular indicators, up to four pending arithmetic operators

The TI-35 Galaxy Solar comes in a carrying case.  The keyboard feels kind of nice. 

Standard Scientific Calculator Feature Set

The TI-35 Galaxy Solar is a landscape calculator with a function set typical of the TI-30 series:

>  Trigonometry:  sine, cosine, tangent, and their inverses

>  Logarithms and powers

>  Polar/Rectangular conversions

>  Decimal/Degrees-Minutes-Seconds (DD.MMSS) conversions

>  Memory Register Operations:  store, recall, sum (STO+), exchange

>  Statistics:  mean (x-bar), standard deviation (σn-1), population deviation (σn)

>  Angle Modes:  degrees, radians, grads, and conversions

The keyboard has an inverse button ( [INV] ).

Inverse Key Mapping

[INV] [SIN]: arcsin

[INV] [COS]: arccos

[INV] [TAN]: arctan

[INV] [EE]: cancel scientific notation display when possible

[INV] [LOG]:  10^x

[INV] [LN]:  e^x

[INV] [P>R]:  R>P (rectangular to polar conversion)

[INV] [DMS>DD]:  DD>DMS (decimal to degree-minutes-seconds conversion)

Rectangular/Polar Conversions

One of the nice things with the TI-35 Galaxy Solar is when these conversions happen, the results are displayed with labels.

Rectangular to Polar Conversion:

Input:  x [ x<>y ] y [INV] [P>R]

Output:  θ [x<>y] r  (with θ and r indicators)

Polar to Rectangular Conversion:

Input:  r [x<>y] θ [P>R]

Output:  y [x<>y] x (with y and x indicators)

Constant Operations

1.  Enter a constant 

2.  Press an arithmetic operation ([+], [-], [×], [÷], [y^x], [INV] [y^x])

3.  Press [INV] (K) 

4.  Enter the number to be operated on then press equals [=].  Repeat step 4 for each number.

Operation Table

a [ + ] [INV] ( K ) n [ = ]...    n + a

a [ - ] [INV] ( K ) n [ = ]...    n - a

a [ × ] [INV] ( K ) n [ = ]...    n × a

a [ ÷ ] [INV] ( K ) n [ = ]...    n ÷ a

a [ y^x ] [INV] ( K ) n [ = ]...    n^a

a [INV] [ y^x ] [INV] ( K ) n [ = ]...    n^(1/a)

Arithmetic Indicators

One of the more unique features of the TI-35 Galaxy Solar are arithmetic indicators of pending operators on the left side of the screen.   The indicators provide an excellent insight to the pending arithmetic operators.

Published are two YouTube shorts that see the arithmetic indicators in action:

Example 1:  https://youtube.com/shorts/GCmuDwjr72w

Example 2:  https://youtube.com/shorts/AOXhkNHI1po

This is my favorite feature of the TI-35 Galaxy Solar calculator.

Final Thoughts

I think this a real good collector's calculator.  I like the landscape form and the pending arithmetic indicators is a great addition.

Eddie 

All original content copyright, © 2011-2023.  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-3650P Program Bank

Casio fx-3650P Program Bank

Economic Order Quantity

The program calculates three business parameters:

*  EOQ:  Economic Order Quantity: this is the optimum number of units to order in a single of order of items to be sold.  Each single inventory item has its on economic order quantity.   The calculation assumes every order will be available when needed.  This number of units is theoretical.  The calculation takes the estimated annual sales of units into account.  

*  The estimated amount of orders needed in a year.

*  The estimated annual cost of inventory.

Note that no amounts are rounded in this algorithm, all results are based on the theoretical values.

Formulas:

EOQ = √(2 × A × B ÷ (C% × D))

where:

A = cost to  place an order

B = annual sales of units

C = holding costs percentage 

D = cost per unit

EOQ is stored in X.

Orders per Year = B ÷ EOQ

The result is stored in Y. 

Estimated Annual Cost = A × amount of orders per year

Estimated annual cost is stored in M.

Program Code:  (47 steps)

(spaces are included for clarity)

Fix 2 : ? → A : ? → B : ? → C : ? → D : 

√ ( 2 A B ÷ 100(^-1) C × D → X ◢

B ÷ X → Y ◢

A Y → M ◢

Norm 1

Example:

Inputs:

A = cost to  place an order = $ 79.36

B = annual sales of units = 100,000

C = holding costs percentage = 24%

D = cost per unit = $ 1.20

Outputs:

X = EOQ ≈ 7,423.69

Y = Orders per Year ≈ 13.47

M = Estimated annual cost ≈ $1,069.01

Source:

Hewlett Packard.   Step-by-Step Solutions For Your HP Calculator: Marketing and Sales:  HP-17B, HP-19B, HP-27S.   Hewlett Packard.  Edition 1.  January 1988. 

Banker's Rounding

We will focus on rounding positive numbers to integers.   

Banker's rounding is similar to regular rounding except, when the number has a fraction part of 0.5, the number is rounded to the nearest even integer.

For example:  both 17.5 and 18.5 would be rounded to 18.  

We can accomplish Banker's rounding by these algorithms, assuming x is a positive number (x>0):

If abs(frac(x)) = 0.5 

Then round(x ÷ 2, 0) × 2

Else round(x, 0) 

IfEnd

If mod(x,1) = 0.5

Then round(x ÷ 2, 0) × 2

Else round(x, 0) 

IfEnd

The fx-3650P does not have a modulus or absolute value function.   We can accomplish absolute value by using √(x^2).

Program Code:  (46 steps)

(spaces are included for clarity)

? → A : Fix 0 : Rnd : Ans → B :

√ ( ( A - B ) ² ) ≠ 0.5 ⇒ Goto 1 :

A ÷ 2 : Rnd : 2 Ans → B :

Lbl 1 : Norm 1: B

Examples:

A = 36.3,  result = 36

A = 36.5,  result = 36

A = 37.5,  result = 38

A = 40.5,  result = 40

A = 41.2,  result = 41

Birthday Probably Function

What are odds that A people/objects do not a share a characteristic in C categories?   

Famously stated:  what are odds that N people do not share a birthday in a 365-day year?  

The probability is calculated as:

p = Π(1 - m ÷ C, m = 1, N - 1) = nPr(C, N) ÷ C^N

Program Code:  (19 steps)

(spaces are included for clarity)

? → A : ? → C :

C [nPr] A ÷ ( C ^ A ) → B

nPr appears as a bold P.

Example:

Probability that 24 people do not share a birthday:

A = 24, C = 365,  result = 0.461655742

Probability that 40 students who do not share a college class, if the college offers 300 classes:

A = 40, C = 300, result = 0.065725193

Voltage Drop of a Copper Wire

The program calculates the voltage drop of a copper wire (conductivity of 12.9).  The wire is assumed to be insulated at 75°C. 

Formula:

1 Phase: (domestic and office appliances)

VD = 2 × 12.9 × B × D ÷ C

3 Phase: (large electronic equipment)

VD = √3 × 12.9 × B × D ÷ C = 1-Phase-VD × √3 ÷ 2

B = current (amps)

C = circular mils* (see below)

D = length of the wire (feet)

C:  

Enter the wire cross area in circular mils.  However, the code allows for three common wire types:

C = 10  (AWG 10):  10,380

C = 12  (AWG 12):  6,530

C = 14  (AWG 14):  4,110

The voltage drop for both 1 phase (stored in X), followed by 3 phase circuits (stored in Y) are displayed.  

Program Code:  (72 steps)

(spaces are included for clarity)

? → B :

? → C :

C = 10 ⇒ 10380 → C :

C = 12 ⇒ 6530 → C :

C = 14 ⇒ 4110 → C :

? → D :

2 × 12.9 × B × D ÷ C → X ◢

X × √3 ÷ 2 → Y

Examples:

Input:

B = Current = 300 A

C = 10  for a #10 AWG wire

D = Length = 24 ft

1 phase voltage drop:  17.89595376

3 phase voltage drop:  15.49835058

Input:

B = Current = 320 A

C = 14  for a #14 AWG wire

D = Length = 30 ft

1 phase voltage drop:  60.26277372

3 phase voltage drop:  52.18909295

Eddie

All original content copyright, © 2011-2023.  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. 

Areas of Right Triangle Knowing the Hypotenuse and the Angle

Areas of Right Triangle Knowing the Hypotenuse and the Angle

General Right Triangle

Let:

D = the length of a hypotenuse

A = the length of side A opposite of angle α°

B = the length of side B opposite of angle β°

Let's assume that we only know of hypotenuse D and angle α°.  Find the area:

Area = 1/2 × A × B

Determined by trigonometric ratios: 

 A = H × sin α° and B = H × cos α°

Then:

Area = 1/2 × A × B

Area = 1/2 × H × sin α° × H × cos α°

Area = 1/2 × H^2 × sin α° × cos α°

Let's assume that we only know the angle β° instead:

Area = 1/2 × H^2 × sin α° × cos α°

Since α° + β° = 90°,

Area = 1/2 × H^2 × sin (90° - β°) × cos (90° - β°)

With the trigonometric identities:

sin(90° - θ°) = cos θ°, and cos(90° - θ°) = sin θ°

Area = 1/2 × H^2 × cos β° × sin β°

In a remarkable conclusion:

Area = 1/2 × H^2 × sin α° × cos α° = 1/2 × H^2 × cos β° × sin β°

Let's look at specific right triangles.

Area of 30°-60°-90° Triangles

Assume that α = 60° and β = 30°.  Then:

Area = 1/2 × H^2 × sin 60° × cos 60° 

Area = 1/2 × H^2 × √3/2 × 1/2

Area = (H^2 × √3) / 8

Similarly,

Area = 1/2 × H^2 × sin 30° × cos 30° 

Area = 1/2 × H^2 × 1/2 × √3/2

Area = (H^2 × √3) / 8

Area of 45°-45°-90° Triangles

On a 45-45-90 triangle, the measures A and B are equal.  Then:

Area = 1/2 × H^2 × sin 45° × cos 45° 

Area = 1/2 × H^2 × √2 / 2 × √2 / 2 

Area = H^2 / 4

Area of 75°-15°-90° Triangles

Assume that α = 75° and β = 15°.  Then:

Area = 1/2 × H^2 × sin 75° × cos 75° 

Area = 1/2 × H^2 × (√6 + √2)/4 × (√6 - √2)/4

Area = 1/32 × H^2 × (√6 + √2) × (√6 - √2)

Area = 1/32 × H^2 × (6 - √6 × √2 + √6 × √2 - 2)

Area = 1/32 × H^2 × 4

Area = H^2 / 8

Similarly,

Area = 1/2 × H^2 × sin 15° × cos 15° 

Area = 1/2 × H^2 × (√6 - √2)/4 × (√6 + √2)/4

Area = 1/32 × H^2 × (6 + √6 × √2 - √6 × √2 - 2)

Area = 1/32 × H^2 × 4

Area = H^2 / 8

Summary

Area of a Right Triangles knowing only the Hypotenuse and One (does not matter which one as it turns out) Angle:

Area = 1/2 × H^2 × sin θ × cos θ

Area of 30°-60°-90° Triangles: (H^2 × √3) / 8

Area of 45°-45°-90° Triangles: H^2 / 4

Area of 75°-15°-90° Triangles: H^2 / 8

Eddie 

All original content copyright, © 2011-2023.  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. 

HP 12C Platinum: Property Value by Income and Reversion Method

HP 12C Platinum: Property Value by Income and Reversion Method

Property Valuation

One way to calculate the value of property is by its income and reversion value. 

In this case, the income is the estimated annual net income before taxes which is assumed to be "paid out" at the end of each year.  This income stream is sometimes known as an Inwood annuity cash flow.

The reversion value of a property is its future value based on factors like inflation, anticipated selling price at the end of the project, and the current market value.  The economic life of the property is not necessarily the remaining physical life of the property.   In a time value of money timeline, the reversion value would be received at the end of the project.

Calculating the value of the property using this income method uses three steps, as illustrated by the Board of Equalization (see the Source section below):

1.  Calculate the value of the income stream, called the net income before taxes (NBIT):

PV_income = NBIT ÷ (SFF + yield% + tax_rate%)

where:

SFF = yield% ÷ ((1 + yield%)^n - 1)

yield% and tax_rate% are expressed in decimal form (Example: 0.05 for 5%)

We can use the TVM keys to calculate ((1 + yield%)^n - 1) by setting the following values:

n,  yield → I/YR%, -1 → PV, 0 → PMT, solve for FV.

Then:  SFF = yield% ÷ FV

2.  Value of the reversion value.

PV_reversion = reversion_value × (1 + (yield% + tax_rate%))^(-n)

We can use the TVM keys to calculate (1 + (yield% + tax_rate%))^(-n)

by setting the following values:

n,  yield + tax_rate → I/YR%, 0 → PMT, -1 → FV, solve for PV.

3.  Calculate the total property value.

Property Value = PV_reversion + PV_income

HP 12C Platinum Program:  Property Value by Reversion Method

Store before running the program:

n =  number of years of remaining economical life of the property

i = yield interest rate

R7 = property tax rate

R8 = net income before taxes (NBIT)

R9 = expected revision value of the property (the value of the property at the end of it's economic life)

Code:

step;  key code;  key

001;  0;  0

002;  44,0;  STO 0

003;  1;  1

004;  16;  CHS

005;  13;  PV

006;  0;  0

007;  14;  PMT

008;  15;  FV

009;  1;  1

010;  30; - 

011;  22; 1/x

012;  45,12;  RCL i

013;  25;  %

014;  45,12;  RCL i

015;  1;  1

016;  25;  %

017;  34;  x<>y

018;  33;  R↓

019;  40;  +

020;  45, 7;  RCL 7

021;  1;  1

022;  25;  %

023;  34;  x<>y

024;  33;  R↓

025;  40;  +

026;  45, 8;  RCL 8

027;  34;  x<>y

028;  10;  ÷

029;  44,40,0;  STO+ 0

030;  45, 12; RCL i

031;  45, 7;  RCL 7

032;  40;  +

033;  12;  i

034;  1;  1

035;  16;  CHS

036;  15;  FV

037;  0;  0

038;  14;  PMT

039;  13;  PV

040;  45, 9;  RCL 9

041;  20;  ×

042;  44,40,0;  STO+ 0

043;  45,12;  RCL i

044;  45,7;  RCL 7

045;  30;  -

046;  12;  RCL i

047;  45,0;  RCL 0

048;  43,33,000;  GTO 000

Examples

Example 1

NBIT:  $7,588  (store in R8)

Reversion Value:  $236,000 (store in R9)

Tax Rate:  1.3%  (store 1.3 in R7)

Annual Rate:  8%  (store in i)

Years of Economical Life Remaining:  5  (store in n)

Property Value:  $180,092.06

Example 2

NBIT:  $12,000

Reversion Value:  $185,000

Tax Rate:  1.6%  

Annual Rate:  7.34%

Years of Economic Life Remaining:  10

Property Value:  $153,286.39

Sources

"Lesson 4 - Time Value of Money (The Income Approach to Value)"  California State Board of Equalization.  Accessed March 27, 2023. https://www.boe.ca.gov/info/iav/lesson4.htm

"Lesson 18 - Property Reversion:  Annuity plus Reversion Method (The Income Approach to Value" California State Board of Equalization.  Accessed March 27, 2023.  https://www.boe.ca.gov/info/iav/lesson18.htm

Eddie

All original content copyright, © 2011-2023.  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. 

Carnival of Mathematics # 216

Carnival of Mathematics # 216

Welcome to a special edition of Eddie's Math and Calculator Blog.  I am once again happy to host the Carnival of Mathematics, which this is the 216th Edition.  The Carnival of Mathematics is facilitated by The Aperiodical.   

Check out the monthly series, where you can also submit articles for future carnival articles, at https://aperiodical.com/carnival-of-mathematics/.

May 2023's Carnival was hosted by Cassandra from Cassandra Lee Yieng's Blog:  https://leeyieng.wordpress.com/2023/05/01/carnival-of-mathematics-215-april-2023/

July 2023's Carnival will be hosted by Vaibhav at Double Root.  Double Root's blog:  https://doubleroot.in/

Gratitude to Ioanna Georgiou for the invitation to host once again.    

The Number 216

In a regular, non-leap year, the 216th day of the calendar is August 4.  On leap years, the 216th day is August 3. 

The digital root of 216 is 9.  (2 + 1 + 6 = 9).

The prime factorization of 216 is 2^3 * 3^3.   Note that 216 = 6^3.  

The divisors of 216 are:  1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, and 216.  The sum of the proper divisors are 384, making 216 an abundant number.  

216 hours is 9 days exactly.

216 can be expressed as sums and differences of powers, including:

216 = 3^5 - 3^3

216 = 2 * 10^2 + 4^2

216 = 12^2 + 9^2 - 3^2

216 = 4^4 - (6^2 + 2^2)

216 = 8^3 - (14^2 + 10^2)

The Carnival

Thank you to everyone who submitted an article.  You can submit entries at https://aperiodical.com/carnival-of-mathematics/.

SineRider 

This is a really cool game about mathematics.  The goal is to set a curve so that the snow-people successfully pass through a designated number of gates.  At first, the curves are linear but they get more challenging as the game goes on.  The game also saves your progress. (Screen shots are from the game.)

View Hack Club's intro video on Youtube:  https://www.youtube.com/watch?v=35nDYoIwiA8

Play the game here:  https://sinerider.com/

The Game of Hex:  Submitted by Massimo Dacasto

The game of Hex has two players.  The goal is to connect a string of hexagons from one side to the opposite side.  Each player claims a hexagon with their own color, such as red and blue.  The game of Hex is featured in the 1980s game show Blockbusters.  

The web page Hall of Hexagons is a gallery of moves a player can make to guarantee themselves a connection over a set number of rows.  Check out the gallery here:  https://www.drking.org.uk/hexagons/hex/templates.html.

Peirce's Law:  Jon Awbrey of the Inquiry Into Inquiry blog

This article explains Pierce's Law and provides the proof of the law.  The proof is provided in two ways:  by reason and graphically.  Simply put, for propositions P and Q, the law states:

P must be true if there exists Q such that the statement "if P then Q" is true.  In symbols:

(( P ⇒ Q) ⇒ P) ⇒ P

Article:  https://inquiryintoinquiry.com/2008/10/06/peirces-law/

The law is an interesting tongue twister to say the least.

Praeclarum Theorema:  Jon Awbrey of the Inquiry Into Inquiry blog

Jon Awbrey provides an explanation and proof of the Praeclarum Theorema, also known as the Splendid Theorem, which states:

For statements a, b, c, and d:  

((a ⇒ b) ^ (d ⇒ c)) ⇒ ((a ^ d) ⇒ (b ^ c))

Article:  https://inquiryintoinquiry.com/2008/10/05/praeclarum-theorema/

A Spiral on the Brighton Seafront:  Sam Hartburn

The place is the Brighton Seafront in Brighton, UK.  On the sea front there is a spiral of poles, which is the ratio of radii of the large circle to small circle is about the Golden Ratio ( (1 + √5) ÷ 2 ).   

See this spiral here:  https://samhartburn.co.uk/sh/maths-tourism-a-spiral-on-brighton-seafront/

Green's Windmill & Science Centre

In another travel related article, Tom, an Education Officer and blog writer of mathematics and astronomy, reviews the Green's Windmill and Science Centre in Nottingham, UK.  By the picture posted on Tom's post, this is a place I would like to visit one day.   Tom praises the Windmill for giving vision to the mathematics behind windmills, including displaying an integral equation related to electricity and magnetism:  

∫ dσ V dU/dw + ∫ dx dy dx V δU = ∫ dδ U dV/dw + ∫ dx dy dz U δV

The blog entry also features the work of George Green, a British mathematician.   

Tom's Blog Entry:  https://tommaths.blogspot.com/2023/05/greens-windmill-science-centre-museum.html

Green's Mill and Science Centre's Web Site:  https://tommaths.blogspot.com/2023/05/greens-windmill-science-centre-museum.html

Golden Integration:  John D. Cook

Speaking of the Golden Ratio, John d. Cook demonstrates Monte Carlo integration but instead of using a purely random sequence, a pseudo-random sequence using the fractional values of n * φ, where φ = ( (1 + √5) ÷ 2 ).  The method is demonstrated using Python.

Article:  https://www.johndcook.com/blog/2023/04/29/golden-integration/

Monte Carlo integration has an advantage of ease of calculation, with a cost of having to sample a large amount of points to get an accurate answer.  

A Plan to Address the World Challenges With Math:  Kevin Harnett of Quanta Magazine

The article is an interview with mathematician Minhyong Kim, Director of the International Centre of Mathematical Sciences (ICMS) in Edinburgh, Scotland.  The goal is ICMS is to focus mathematics with humanitarian studies. 

In the article, Kim emphasizes that young mathematicians want to work in diverse fields outside the traditional subjects.  

Article:  https://www.quantamagazine.org/a-plan-to-address-the-worlds-challenges-with-math-20230511/

Web page for ICMS:  https://www.icms.org.uk/

The ICMS's web page for The Mathematics for Humanity program, which has recently launched:  https://www.icms.org.uk/funding-opportunities/mathematics-humanity

Hamilton's Quaternions, or, The Trouble with Triples:  Article by James Propp

Math with a romantic story between William Rowan Hamilton and Catherine Barlow.  Hamilton's infatuation for Barlow lasted, despite the fact that Barlow was married to another man.   Barlow's son needed a tutor in mathematics, and Hamilton was the man to call.  While Hamilton was tutoring her son, he and Barlow discussed Barlow's unhappy marriage.   

Propp then details who Hamilton and Barlow first met, along with the math of quaternions in detail.  

A quaternion is defined an extended complex number represented in the form

q1 * i + q2 * j + q3 * k + q4, with q1, q2, q3, and q4 are real numbers.

When multiplying quaternions the following rules are observed:

i^2 = j^2 = k^2 = -1

i * j = k,   j * i = -k

j * k = i,   k * j = -i

k * i = j,   i * k = -j

The article:  https://mathenchant.wordpress.com/2023/05/17/hamiltons-quaternions-or-the-trouble-with-triples/

Degree of ζ(3)

Here is a stack exchange talking about the integral representation of ζ(3)  (Riemann Zeta function of 3).  As of June 2, 2023, the post has the option question regarding a degree of a group of permutations.   

The open question:  https://math.stackexchange.com/questions/4703662/degree-of-zeta3-as-a-period

I would like to thank everyone who submitted articles for the Carnival of Mathematics.  Gratitude to the Aperiodical for having me as this month's host.   

Eddie 

All original content copyright, © 2011-2023.  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.