1/2 Rho
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Not quite that simple my friend.
The relationship of density to temperature and pressure can be expressed as... P (pressure in mb) divided by [T (absolute temperature) x Density (Rho)] = Constant. Where Rho is mass per unit volume (usually expressed as kg per m3).
When air is compressed, a greater amount can occupy a given volume; i.e. the mass, and therefore, the density, has increased. Conversely, when air is expanded less mass occupies the original volume and the density decreases. From the above formula it can be seen that, provided the temperature remains constant, density is directly proportional to pressure, i.e. if the pressure is halved, so is the density, and vice versa.
Hope this helps.
Oooops... I should have read the question. Genghis is obviously on the ball.
The relationship of density to temperature and pressure can be expressed as... P (pressure in mb) divided by [T (absolute temperature) x Density (Rho)] = Constant. Where Rho is mass per unit volume (usually expressed as kg per m3).
When air is compressed, a greater amount can occupy a given volume; i.e. the mass, and therefore, the density, has increased. Conversely, when air is expanded less mass occupies the original volume and the density decreases. From the above formula it can be seen that, provided the temperature remains constant, density is directly proportional to pressure, i.e. if the pressure is halved, so is the density, and vice versa.
Hope this helps.
Oooops... I should have read the question. Genghis is obviously on the ball.
Last edited by TheChitterneFlyer; 4th Jan 2008 at 10:25.
I think the question was about the ½ wasn't it?
The equation is derived from the formula for kinetic energy, which is ½MV², from which the value for dynamic pressure: ½Rho.V² is reasonably simple.
If you look in pretty much any aerodynamics textbook (the grown up ones, not standard pilot training ones) you'll find the derivation of Bernoulli's formula, from which it's reasonably clear how this happens.
G
The equation is derived from the formula for kinetic energy, which is ½MV², from which the value for dynamic pressure: ½Rho.V² is reasonably simple.
If you look in pretty much any aerodynamics textbook (the grown up ones, not standard pilot training ones) you'll find the derivation of Bernoulli's formula, from which it's reasonably clear how this happens.
G
Didn`t you start on triplanes,Jenks ?
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The answer lies in aviation history. In the earliest days of Central Flying School, all flying machines were biplanes, and the figure was "Rho." The change to "Half Rho" came about with the advent in great numbers of the monoplane.
B
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1/2 because the (1/2rho*V^2) component of the lift formula is the mathematic indefinite integral of rho*V .
It is also mathematically convenient for it to be 1/2 in the lift equation because the quantity (1/2rho*V^2) can be calculated directly from Bernoulli's equation if you know the pressure.
All of this theory applies only to inviscid flows however, once you have viscosity in the frame it's a whole different (and more complex) ball game!
Also, applying Bernoulli's equation to wing cross-sections can be quite tricky and requires a mathematical concept known as a conformal transformation to make the geometry (if not the algebra) somewhat easier to deal with!
B
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JENKINS. Speaking of the question's relationship to aviation history, surely you were referring to the 1909 Tri-plane of Mr A V Rho?
Last edited by GOLF_BRAVO_ZULU; 4th Jan 2008 at 12:08. Reason: Finger Trouble
