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Old Jan 31, 2003, 12:55 PM
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Toronto,Canada
Joined Jul 2002
435 Posts
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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Old Feb 02, 2003, 11:52 AM
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Dick Huang's Avatar
Dallas,TX, USA
Joined Mar 1999
834 Posts
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
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Old Feb 02, 2003, 01:47 PM
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Sparky Paul's Avatar
Palmdale, CA
Joined Oct 2000
13,371 Posts
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.
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Last edited by Sparky Paul; Feb 02, 2003 at 01:50 PM.
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Old Feb 02, 2003, 05:00 PM
Registered User
Toronto,Canada
Joined Jul 2002
435 Posts
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
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Old Feb 03, 2003, 07:39 PM
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Cincinnati, OH
Joined Aug 2002
767 Posts
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.
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Old Feb 04, 2003, 01:12 AM
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Toronto,Canada
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.
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Old Feb 04, 2003, 01:13 AM
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Toronto,Canada
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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.
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Old Feb 04, 2003, 05:00 AM
Senior Moment Member
Virginia Beach, Va.
Joined Jan 2002
1,233 Posts
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
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Old Feb 04, 2003, 10:37 PM
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Ridgecrest, CA
Joined Feb 2003
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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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