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Mach TAS and SAT

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Old 1st February 2005 | 21:08
  #1 (permalink)  
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From: luxembourg
Devil Mach TAS and SAT

HI

As far as I can remember for a given Mach number and a Static air temperature there is a unique True air speed.

How are they related?

I will need and equation which should look like this:

TAS=M*constant*SAT

I can also handle square roots and fractions.

Many thanks
fernando is offline  
Old 1st February 2005 | 22:14
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From: various places .....
If you want to strain the grey matter a bit, refer to any standard undergrad aerodynamics text .. in essence, the story is

M = TAS/local speed of sound

where local speed of sound, a

a/a0 = SQRT (T/T0)

where subscript 0 indicates standard sea level values and T is absolute (Kelvin or Rankine),

eg a0 = 661.48 kt and T0 = 288.15 K = 15 C

Rearranging for TAS

= M * a

= M * a0/SQRT(T0) * SQRT(T)

Substitute for whatever system of units is your cup of tea .. add a few unit conversion factors to taste and you end up with the sort of equation you are looking for .. something along the lines of

For

TAS kt
OAT C

= M * 661.48 / SQRT(288.15) * SQRT(273.15 + C)

= 38.97 * M * SQRT(273.15 + C)

(assuming I haven't made any numeric errors along the way .. )
john_tullamarine is offline  
Old 2nd February 2005 | 20:31
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From: luxembourg
many thanks

I got it. thank you.
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Old 4th February 2005 | 09:15
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From: france
hello every one

tas= 661.M.sqrt theta with theta=T\Ts T=(273+sat°c )K Kelvin & Ts=288K(273+15°c)

to find this formula:

tas=M.a & a square=k.R.T k=1.4 for dry air =cte
R(universal gas cte)=287joules\kg & K
T=temperature in kelvin

now take above formula & multiply & divide at the same time by a0=speed of sound at sea level=661kts.

so: tas=ao.M.squareroot of k.R.T\squareroot of k.R.Ts
a0square being of course=k.R.Ts
theta=k.R.T\k.R.Ts=T\Ts

so finally: tas=661.M.sqrt theta

equally eas=661.M.sqrt delta with delta=p\po p=atm pressure in Hpa & po=1013.25Hpa

delta=sigma x theta ; sigma=rho(density in kg\m3)\rhozero
rhozero=1.225kg\m3

to find above formulas think of mr boyle's gas laws: p=rho.R.T & po=rhozero.R.Ts divide both formulas & find: p\po=rho\rhozero x T\Ts

thus: delta=sigma x theta

conclusion: tas=function of mach & temperature whereas ias & eas are function of mach & pressure

so we see it every day in climb:
first climb at cte eas, mach increases & theta decreases
then at crossover altitude climb at cte mach & eas & delta now decreases.

stop here before my headache starts again
blackmail is offline  
Old 4th February 2005 | 18:06
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LGB
 
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From: -
I can confirm

LSS = 38.95 x SQR (temp Kelvin)

or

LSS = 38.95 x SQR (temp C + 273)

LSS = Local speed of sound (KTAS). Multiply by desired Mach number to get the appropriate TAS.

I can recommend "Air Data Conversion" programme, should be available on http://www.avionic-cbt.co.uk/adc/adcsetup.exe but it does not work right now? It is a 30-day trial version.

Edited for an "oops" ...
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Old 5th February 2005 | 05:05
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From: Australia
LGB,

I think you'll get slightly better results if you use a Constant of 38.97, and use +273.15 for your conversion to Absolute Temperature.

Oh Gawd ! I'm becoming too pedantic.

Regards,

Old Smokey
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Old 7th February 2005 | 00:19
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From: Canada
Wow, I'm really impressed with the scientific knowledge of some of the members to this website. I knew the answer to the initial question was something like 39 * sqrt(temp (K)), but I had no idea some of you people could readily derive the equation. I have an engineering degree, so I could follow the logic of the responses, but I didn't know the original form of the equation. Kudos!
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Old 7th February 2005 | 02:15
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From: various places .....
.. stock standard undergrad aero eng textbook stuff ..
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