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HP 32S and HP 32SII Week: Total Drag

HP 32S and HP 32SII Week:  Total Drag





Introduction


The following equation calculates the total drag force applied to parallel to an area in the opposite direction of the object's motion.  


Fd = 1/2 * ρ * v^2 * Cd * A,   ρ = P / (R * T)


ρ = air pressure 

P = absolute pressure (101,325 Pa, 14.696 psi, 2116.224 lb/ft^2)

R = specific gas constant (287.03 J/(kg k), 53.3533 (ft lbf/lb °R))

T = temperature (K = °C + 273.15, °R = °F + 459.67)


A = cross area that the drag force is applied (m^2, ft^2)

V = velocity of the object (m/s, ft/s)

Cd = drag coefficient (unitless)


The program gives outputs:


ρ = air pressure (kg/m^3, lb/ft^3)

Fd = total drag (kg*m/s^2, lb*ft/s^2)


HP 32S/32SII Program:  Total Drag, SI Units

Size: 55.5 bytes


F01 LBL F

F02 INPUT C

F03 INPUT A

F04 INPUT V

F05 INPUT T

F06 273.15

F07 +

F08 287.03

F09 ×

F10 1/x

F11 101,325

F12 ×

F13 STOP

F14 RCL× C

F15 RCL× A

F16 RCL V

F17 x^2

F18 ×

F19 2

F20 ÷

F21 STOP


Example:

C = drag coefficient = 0.31

A = area =7.0686 m^2

T = temperature = 18.8 °C

V = velocity = 3 m/s


Results:

ρ = 1.20915184207 kg/m^3

Fd = 11.9230799416 kg*m/s^2


HP 32S/32SII Program:  Total Drag, US Units

Size: 55.5 bytes


F01 LBL F

F02 INPUT C

F03 INPUT A

F04 INPUT V

F05 INPUT T

F06 459.67

F07 +

F08 53.3533

F09 ×

F10 1/x

F11 2116.224

F12 ×

F13 STOP

F14 RCL× C

F15 RCL× A

F16 RCL V

F17 x^2

F18 ×

F19 2

F20 ÷

F21 STOP


Example:

C = drag coefficient = 0.31

A = area =76.0868 ft^2

T = temperature = 65.84 °F

V = velocity = 9.84252 ft/s


Results:

ρ = 7.54778228E-2 lb/ft^3

Fd = 86.232900381 lb*ft/s^2



Source:


Lindeburg, Michael R. PE   Civil Engineering Reference Manual for the PE Exam 14th Edition  Professional Publications, Inc:  Belmont, CA.  pp. 17-41 and 17-42


All original content copyright, © 2011-2022.  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 32S and HP 32SII Week: Hypergeometric Distribution

 HP 32S and HP 32SII Week:  Hypergeometric Distribution





What are the Odds?


The hypergeometric probably function deals with taking samples without replacement.  The trials are not independent.  The probability formula is:


G(S; N, T, M) =

1 ÷ nCr(N, M) * nCr(T, S) * nCr(N-T, M-S)


S = number of successes

N = main population size

T = target's population size

M = sample size


nCr(x, y) = x! ÷ ( y! * (x - y)!)


HP 32S/32SII Program:  Hypergeometric Distribution Probability

Size: 30 bytes


H01 LBL H

H02 INPUT S

H03 INPUT N

H04 INPUT T

H05 INPUT M

H06 RCL N

H07 RCL M

H08 Cn,r

H09 1/x

H10 RCL T

H11 RCL S

H12 Cn,r

H13 ×

H14 RCL N

H15 RCL- T

H16 RCL M

H17 RCL-S

H18 Cn,r

H19 ×

H20 STOP


Examples:


What are the odds that four hearts are dealt out of a five card hand?  Assume a standard, 52 card deck.


S = 4 (4 hearts)

N = 52 (52 cards)

T = 13 (13 hearts)

M = 5 (5 card hand)


Result:  1.07292917E-2 (≈1.07%)


What are the odds that a pair of Kings are dealt out of a five card hand?  Assume a standard, 52 card deck.


S = 2 (2 Kings)

N = 52 (52 cards)

T = 13 (13 hearts)

M = 5 (5 card hand)


Result:  3.99298181E-2 (≈3.99%)


Source:


"Hypergeometric Distribution"  Texas Instruments Programmable Slide-Rule SR-56 Applications Library  pp. 58-59 Texas Instruments, 1976


Download the document here, with gratitude to Datamath:

http://www.datamath.net/Manuals/SR-56_AL_US.pdf


Eddie



All original content copyright, © 2011-2022.  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. 


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