Swept Volume Theory

First you nit-pick about infinitesimal differences that are negligible for engineering purposes. Then, you claim that air is "equally compressible" at different velocities as if it was an ideal gas.

 

Below is an excerpt from a translation of Bernoulli's Book Hydrodynamica
While almost all of Bernoulli's experiments involved water, he discusses "elastic fluids" compressible by our terminology , and recognizes that the faster moving smoke in a stack can draw in (obviously by the lower pressure within the stack)
ambient pressure air through a hole through the stack. The translator refers to the fact that Bernoulli did not recognize streamlines which Euler (discovered, might be a poor term) included/referenced in his work.
Bernoulli also included a quite lengthy chapter on "elastic fluids" (compressible) but not of interest to us in the topic of this thread.
I only included this as Sailor Al stated that Bernoulli's work only dealt with liquids.
Of course in 1740 ish, water poured into vessels and subsequent measurement were pretty much all that Bernoulli had to work with when doing his experiments at that time, almost 300 years ago.

I was surprised to see the Integral sign in Bernoulli's book as I assumed that this was a much later discovery but another mathematician had come up with this 75 odd years earlier and referenced another Bernoulli

"His results on the integral calculus were published in 1684 and 1686 under the name 'calculus summatorius', the name integral calculus was suggested by Jacob Bernoulli in 1690."

§. 17. Again there are other natural phenomena, the explanation of which truly depends on this static-hydraulic theory: such as how smoke rising through a chimney may draw air through an opening made in the chimney after itself with great impetus ; how a wind blowing from a more confined place into one more open may lose some of its elasticity, just as may be gathered from that, as open windows may be closed by the air, by that escaping from the room on account of retaining its greater elasticity ; and others of this kind, which one cannot examine individually. The pressures of moving fluids indeed are able to be varied in innumerable ways ; yet I think all can be reduced to our general principle : we will examine two examples of this theory ; the first I have deduced from the known motion, which a fluid is going to have, if in the place of determining the pressure the vessel may be perforated with an infinitely small hole: the other, as they say, to be deduced a priori from our general theory ; and often each may be obtained in the same place, as one may require the help of the other, and then for the other an estimation of the pressure arises, as I will show be a single example. §. 18. We may consider in the vessel, as set up in Figure 72, the horizontal tube to have a

 

To be continued? I can't wait for the next chapter.

 

Milne-Thomson and Anderson both use very long, complex and totally different arguments to explain why, in the mathematics underlying aerodynamics (the subject of their books), air can be considered incompressible below 300 mph (Milne-Thomson), or Mach 0.3 (Anderson). Neither of them claim that air is incompressible, indeed Anderson freely admits:



It seems to me that if you can use a logical argument to prove a myth, then there has to be a problem with the argument.
That's what I find hard to accept.

 

There is no problem with the statement. In engineering, infinitesimal values can be ignored. At the end, engineering is based in statistical values, not exact ones. For example, there is no 100% accurate measurements. They all have some +/- tolerances.

 

Milne-Thomson's argument takes less than two pages. You consider that long & involved?

 

Come on! The argument on pages 34 and 35 relies on equations developed over pages 29 to 33 .

That's seven pages.

Yup!

 

For the sake of argument assume air is fully compressible in all conditions.
Your Swept Volume Theory is based on the analysis of how sails work.
A good sailing breeze say 15 MPH produces a pressure of 0.00256 x 15 x 15 = .576 psf
Convert to psi .576 / 144 = .004 psi
Atmospheric pressure 14.7 psi
Percentage change in pressure .004 / 14.7 x 100 = .02721
How is such a trivial change of pressure and the resulting change in volume relevant?

 

I hate to quibble, but your units aren't consistent. You haven't converted the speed from mph to ft/sec.

 

@latestarter
Remember that it only takes about 5% ATM to keep a 747 in the air and around .5% ATM for a little Cessna.
How much pressure do you think a yacht needs?

 

Why does it take 1o times more to keep a 747 than a Cesssna?

 

A 747 is about 500 times heavier than a Cessna but it has a wing area about 50 times larger. 500/50 = 10.
Pressure = force/area.

 

My understanding is that the co-efficient 0.00256 includes the conversion.
I used this formula on Wikihow:-

"3 Calculate wind pressure. The simple formula for wind pressure P in imperial units (pounds per square foot) is P=0.00256V2 https://wikimedia.org/api/rest_v1/media/math/render/svg/d52273a1c31dd3def5c2f0872ad41043a0244292, where V is the speed of the wind in miles per hour (mph).[4] To find the pressure in SI units (Newtons per square meter), instead use P=0.613V2 https://wikimedia.org/api/rest_v1/media/math/render/svg/b9a7b43953ea57a2f33313c77a5950f0dd23897c, and measure V in meters per second.[5]

• This formula is based on the American Society of Civil Engineers code. The 0.00256 coefficient is the result of a calculation based on typical values for air density and gravitational acceleration.[6]
• Engineers use a more accurate formula to take into account factor such as the surrounding terrain and type of construction. You can look up one formula in ASCE code 7-05, or use the UBC formula below.
• If you're not sure what the wind speed is, look up the peak wind speed in your area using the Electronic Industries Alliance (EIA) standard. For example, most of the U.S. is in Zone A with 86.6 mph wind, but coastal areas might lie in Zone B (100 mph) or Zone C (111.8 mph)."

4 Ways to Calculate Wind Load - wikiHow https://www.wikihow.com/Calculate-Wind-Load

Sailor Al wrote
"Remember that it only takes about 5% ATM to keep a 747 in the air and around .5% ATM for a little Cessna.
How much pressure do you think a yacht needs?"

Very little in comparison to atmospheric pressure, which is the point I was making.
A very small change in pressure will produce a very small change in volume which can be ignored for now, as you are yet to quantify your theory.

It is unfortunate but probably inevitable that what started as a discussion on sails morphed into aircraft as they introduce unnecessary complexities.

 

Basically your formula is just a simplification for the dynamic pressure (0.5*rho*v^2) where the air density is already inserted (1.225 kg/m^3) 0.5*1.225 ~ 0.613

 

Sorry, what tripped me up was that I recognized 0.00256 as the density of air in slugs/ft^3 and didn't bother looking at the rest of the terms.
In my defense, I've got the flu (although that's not really much of a defense)