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Jan 31, 2003, 12:55 PM
Registered User

airspeed vs dynamic air pressure


Given "standard day" conditions, what is the relationship between impact pressure on a pitot tube and the indicated airspeed ?

I found this formuala :

velocity = square root of (2 x [impact pressure - static pressure] / air density).


To solve for Vel = 1, (1 what ?? .. ft/sec ?), I used Static= 14.7 lb/in^2, air_density= .00237 slug/ft^3

got impact_press = 14.701185 lb/in^2

Is that correct ? (14.701185 lb/in^2 pressure difference = 1 ft/sec ?


If so, how would I solve for Vel=XX. I realize it's a "basic" math question, but basic math is not one of my strong points

Mike
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Feb 02, 2003, 11:52 AM
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Dick Huang's Avatar
Hi Mike,
I use the equation for dynamic pressure(q) where
q= 1/2 Rho*V^2
q is in lbs/ft^2, Rho is in slugs/ft^3 or lbs-sec^2/ft^4 and V is in ft/sec.

If we assume 88 ft/sec (60 mph) for V,
then q=0.5*0.002377*(88)^2=9.20 lbs/ft^2. which will be the total pressure on a pitot tube.

Dick Huang
Feb 02, 2003, 01:47 PM
Ascended Master
Sparky Paul's Avatar
Total pressure is static + dynamic (q).
A conventional airspeed indicator subtracts (it's vented to static) static from total to show "airspeed", which is actually the dynamic pressure. The bent tube that runs the needle in the airspeed indicator reads total directly.
.
The altimeter shows static pressure.
Last edited by Sparky Paul; Feb 02, 2003 at 01:50 PM.
Feb 02, 2003, 05:00 PM
Registered User
Quote:
Originally posted by Dick Huang
Hi Mike,
I use the equation for dynamic pressure(q) where
q= 1/2 Rho*V^2
q is in lbs/ft^2, Rho is in slugs/ft^3 or lbs-sec^2/ft^4 and V is in ft/sec.

If we assume 88 ft/sec (60 mph) for V,
then q=0.5*0.002377*(88)^2=9.20 lbs/ft^2. which will be the total pressure on a pitot tube.

Dick Huang

Thanks Dick.

I've read about "q" before, time to review it again

Mike
Feb 03, 2003, 07:39 PM
Registered User
QUOTE]Given "standard day" conditions, what is the relationship between impact pressure on a pitot tube and the indicated airspeed ? [/QUOTE]

An interesting discussion, but why do you ask?

In terms of modeling, the important thing to know is "q", calculated as Dick Huang suggests. Seeing that q is a function of the velocity squared, in terms of Standard day, 20 fps (13.6 mph) equates to about 7.6 oz/sq ft, 40 fps to 30.4 oz/sq ft, and and 60 fps to 68.5 oz/sq ft. These are the numbers that relate to stall speeds for a given wing loading, design CL, etc.
Feb 04, 2003, 01:12 AM
Registered User
.
Feb 04, 2003, 01:13 AM
Registered User
Quote:


An interesting discussion, but why do you ask?

Why is the sky blue ?

Nothing really to do with *practical* modeling (airspeed question).

But I figured someone here would know the answer.
Feb 04, 2003, 05:00 AM
Senior Moment Member
To add a little more to it, 'indicated airspeed' on the dial is not true airspeed. True airspeed is indicated airspeed adjusted for temperature (density). Full scale pilots use indicated airspeed plus outside air temp to compute their true airspeed.

Jimmy
Feb 04, 2003, 10:37 PM
Registered User
just to add a little confusion--don't forget to add calibrated airspeed (indicated airspeed corrected for position error) between indicated and true.


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