Huh Yea...lets dont let this get out of control. I will put together a parts list. But just to be sure the main purpose wasa to air down fast and have the same pressure.

And to prove it works.

A = pi * r^2 *4 (1 airhose * 4 chucks) First of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and

BAF = BAC + CAF = CAB + BAE = CAE.
ABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.

Similarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.

The square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get

c2 = (a-b)2+2ab
= a2-2ab+b2+2ab
= a2+b2

This proof is ascribed to Leonardo da Vinci (1452-1519) [Eves]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle ABH is 45o. This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45o.) Now, area(ABHI)+area(JHBC)=area(ADGC)+area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem.

The math is here guys so no need to argue the math facts.

So Yea this proves it for sure...it is way too much faster. later...Clint