Air Gun Home :: View topic

rsterne · Moderator

I found an interesting graph on Transonic Drag that I thought was worth sharing with you....

Note that the drag starts to rapidly increase between Mach 0.80-0.85 (900-950 fps) and has doubled by Mach 0.9 (1000 fps) and tripled by Mach 0.95 (1050 fps).... This means that the extra energy it takes to get a pellet above 950 fps will quickly be shed in the first few yards anyway.... That ties in exactly with the work I have been doing on how the Ballistics Coefficient changes as you push the velocity close to the speed of sound....

Since 950 fps gives you an energy of 2 FPE per grain of pellet weight (an easy number to remember).... this is just one more reason not to push the velocity past that....

Bob

Big Ern · Veteran Member

Looks like you were right when you said the magic number was 900 Bob.

Slavia · Moderator

You can certainly see why they once called it the "sound barrier" in aviation - until new technology came around it was an impenetrable wall.

But you really have to look at the supersonic part of the graph, where drag comes back down. I think Bob is prepping us for a Mach 2.5 Disco project.

rsterne · Moderator

Ah, but the drag doesn't come back down, just the CD.... If the CD were a constant, the drag would be going up by the square of the velocity.... To find the relative drag force, you would have to multiply the velocity squared by the CD, so in terms of the drag, you would get numbers more like the following:

Let the drag at Mach 0.7 be "10 units" (we have to have some value) times the CD of 0.3 = 3.0 "units"
At Mach 0.8, the drag due to velocity would be (0.8/0.7)^2*10 = 13 times the CD of 0.33 = 4.3 "units"
At Mach 0.9, the drag due to velocity would be (0.9/0.7)^2*10 = 17 times the CD of 0.6 = 10 "units"
At Mach 1, the drag due to velocity would be (1/0.7)^2*10 = 20 times the CD of 1.2 = 24 "units"
At Mach 1.5 the drag due to velocity would be (1.5/0.7)^2*10 = 46 times the CD of 0.68 = 31 "units"
At Mach 2, the drag due to the velocity would be (2/0.7)^2*10 = 82 times the CD of 0.43 = 35 "units"
At Mach 2.5, the drag due to the velocity would be (2.5/0.7)^2*10 = 128 times the CD of 0.4 = 51 "units"

There is 8 times the drag force at Mach 1 than at Mach 0.7.... Then after that, the drag builds more slowly than the "velocity squared" rule would indicate (because the CD is decreasing).... At least that's what I get out of the graph.... If we look at the RELATIVE drag in the velocity range that concerns us, we get the following:

Mach 0.7 (791 fps) drag is "X"
Mach 0.8 (904 fps) drag is 1.4 X
Mach 0.9 (1017 fps) drag is 3.3 X
Mach 1.0 (1130 fps) drag is 8.0 X

This information, combined with the actual measurements I have made of Ballistics Coefficients at varying velocities, is enough to convince me that pushing past about 950 fps is, quite literally, a drag....

toadmyster223 · Senior Member

Does this figure apply directly to projectiles of all imaginable calibers, or just one?

Alstone · Moderator & Site Supporter

I’m really glad you explained all that bob it makes me feel really inferior with a maximum of 12 fpe. I shall just have to dream. Shocked

Nice work! Thumb Up!

Al

rsterne · Moderator

AFAIK, the graph is "generic" and not specific to one given projectile.... It is similar to other graphs I have seen regarding why transonic flight takes huge amounts of power, but once you reach Mach 1.5, then Mach 2 isn't that much harder to achieve....

Bob

jedi · Member

Just seen this......Excellent bit of information.

Reminds me of this episode of mythbusters for some reason....

Supersonic bullets disintergrate within 2-3 yards in water where as the subsonic ones travel thurther and still have impact power...

cant find the full episode, but they try a range of armory

http://www.youtube.com/watch?v=yvSTuLIjRm8

Full episode here I belive

http://www.vbox7.com/play:5d66bc76