R/C Naval Combat

Hi John, I just made a suggestion, I'm not using that design cannon so I'm not doing the heavy lifting :)

quote:

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Originally posted by JustinScott

AAAHH.... Piss off! :) [:0]

I'm going to go sob alone in my corner & play with my friends... (all me of them.) [V]


[:D][;)][:o)]

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UUmmm... You didn't think I was hacking on you... did you??

After I re-read the post it may have seemed that way, but I can assure you that is not the case!!.... Unfourtunatly, text dosn't always do the best job of portraying emotion..... but the smileys help[:D]

Anyhow, if I offended anyone, it was certainly not my intention.
Just havin' fun[:p]


That being the case:

quote:

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Originally posted by Tugboat


Is the gorilla african or european? oops, that's swallows.

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This is too funny. I'm now considering arming my ship with the holy hand grenade. Maybe I can get the gorilla to throw it.... if I can just figure out how high to have him to count!![;)]

quote:

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Originally posted by JustinScott

The site doesn't say... However, isn't it just a little pneumatic pushrod? So wouldn't it be the same pressure as the CO2 behind it? So 150psi? Which is why the accumulator pressure is usually a little lower, so there is a pressure difference?

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In a way it would be the same as the air pressure behind it...

the psi is the operative term...

if we assume a piston, a rod, or something similar, with a surface area of 1 square inch the pressure exerted on it @ 150psi would be 150 lbs, therefore it would push with 150 lbs of force against whatever would try to restrain it. However, a piston with a surface area of 2 square inches would exert 300 lbs of force at the same 150psi.

Thanks to Tugboat, we know the MPA-7 has a bore (therefore a "piston") of 7/8" dia.

So if we apply the formula to calculate the area of a circle (pi*radius squared)

radius of 7/8=.4375

.4375*.4375=.1914

pi of course = 3.1415

.1914*3.1415=.6013

therefore the surface area of a 7/8 piston = approx .6 sq in.

.6*150 pounds per square inch = 90 lbs of force

if we apply the same formula to a different sized piston, say 2"

1*1=1

1*3.1415=3.1415

a 2" dia. piston has a surface area of 3.14 square inches

3.14 * 150 pounds per square inch = 471 lbs of force


So assuming the same pressure on both the MPA-7 and the buna valve (it may or may not be the same, but for comparison it's easier to assume that it is) all that needs to happen is the MPA-7 piston has to have a larger effective dia. than the buna ball. What the effective diameter of the ball is I don't know, due to the curvature of the ball, and how much of it is exposed to the air pressure in the accumulator, due to the size of the seat it sits on?? Any mechanical engineers out there know how to figure it??

The effective surface area of the buna ball is the through-hole of the valve. Any part of the ball that extends beyond the hole will be canceled out.

crude diagram:

1....-1
0....-1
0....-1
0....-1
1....-1

The top and bottom row are the parts of the ball that extend beyond the hole. The second, third, and fourth rows are the hole. 1 represents force being applied in one direction, -1 represents force being applied in the opposite direction, 0 represents no force being applied. Total up the force on one side and compare it to the other side. You will find that the sum (the net force) is equal to the force over the hole. That is the force that the actuator must overcome.

OK, so I'm not going to read through all those calcs tonight... I've got work tomorrow. However, I thought I'd post real quick.

It is nearly impossible for you to offend me, unless you are trying. I'm just here to BS & have fun. :)

So based on what Gascan is saying, basically the red arrows in the image cancel each other out, and we are effectivley left with what Kotori was saying (smart guys, those two!), the aperture diameter.

Based on an aperture dia. of .500" (from the JC White drawings -- actual conditions may vary!)

Run the numbers and we get around 29.5 lbs of force required to open the ball. Of course this is based on an even pressure, as Justin said, impact force would probably dramaticly reduce this figure, if the piston had some "free stroke" and a little weight to it, or as Tugboat said, the cannons do not require 150 lbs of pressure to penetrate the foam[:D]

Seem right to you guys??


Ummm.... did I hijack this thread? Sorry, the discussion just got interesting!![:)]

Exactly right, Cannonman. Nice diagram! Beats the heck out of mine. .500 inches is the most common aperture I know of. The torpedoes my brother and I made used a larger aperture to ensure extremely rapid evacuation of the accumulator and give the most powerful thwonk possible (yes, I'm using the word "thwonk" in a technical discussion). 5/8" comes to mind, but I need to check with Carl to be sure.

Did the larger aperture supply a noticibly more powerful "thwonk"?

I suppose it would be more likely to on the torpedos, but I'm wondering if the restriction where the air must flow thru the adapter on an Indiana style cannon would possibly cancel out some of the "thwonk" gained with the larger ball valve.

When I get a chance I'll run some numbers on to see if the ball valve is a bigger restriction, or the bearing adapter. I'm sure a bigger valve would be better regardless, provided space is not an issue.


P.S. Is thwonk in the dictonary? If not, maybe we could get Websters to add it, seems like quite a useful word!![^]