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Arrg! me gots a question.

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Musical_Magic  
1 Jul 2009 23:38 | Quote
Joined: 29 May 2008
United States
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Dont ask about the pirate intro, i really dont know...Anyways, this is an importiant question.

On your strat, ect tremolo springs how much pounds is each one pulling? (at the standard length between the tremmolo main piece and the tremolo anchors)

and two, I'm not using this for anything guitar related, i'm acctualy using the springs as a very powerful launcher for a toy glider i made. is there anyway i can convert the pounds its pulling to what speed the plane will be going when launched? (I do plan to add weight and other stuff,so its not going to just be pounds = speed) And i have a Bad feeling this is mathematical, so if its above 8th grade level...Explain :P. And of course, if what i'm talking about isent possible...Lower me to somthing...Practicle


And whoever answers this needs to give themselfs a pat on the back for good memory XD (unless they're taking this stuff in school/collage atm)


Thank you (even though i really should wait to say that XD, oh well, its polite being i'm giving ya'll really nothing...exept answering questons, even though i'm most of the time on the same level as the guy ASKING the question)

*Deep breath*

That was a mouthfull
Guitarslinger124  
2 Jul 2009 05:18 | Quote
Joined: 25 Jul 2007
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Usually for a strat if you're using .10 gauge strings you'll have between 16 and 18 pounds of pressure from the high E string to the Low E string. Obviously the High string with have less pressure. So essentially your tremelo wont dip equally on both sides, but that can be adjusted.

I'm not sure if I can think of an equation right off the top of my head, but here are some things to take into account: Weight of your plane vs pounds of pressure on the springs, length of launch pad and the aerodynamics of your plane.
An indirect way to figure this out would be to first simply find out how fast your plane went overall which you can do using this formula:
SPEED (F/S) = DESTANCE TRAVELED (F) / TIME (S) where F = Feet and S = seconds.
So for example lets say you launch your plane from a spring that is producing 16 pounds of pressure. If your planes travels 30 feet in 15 seconds then your speed would be 2 feet per second. Which means 16 pounds of pressure will make your plane travel 2 feet per second (which by the way, is 1.36 miles per hour).
I'll think about it some more and get back to you.
Good luck dude.
Guitarslinger124  
2 Jul 2009 13:15 | Quote
Joined: 25 Jul 2007
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Just an update on what I have so far. I just made up numbers because you didn't give me any. I didnt realize how rusty I was hehe...

Here is what I have so far based on a few Equations. F=ma, F=-kx and U=1/2*kx^2 (That's speed, spring constant and potential energy).
F = -kx
F = mg = (0.500kg)(10m/s^2) = 5N.
x = 45cm - 30cm = 15cm = .15m
k = 5 N / 0.15 m = 33 N/m.

U = 1/2 kx^2
U = .5 * 33N/m * .15m^2

I'm not solid on the answer yet, I messed up somewhere with my conversions. Too bad I don't have a scanner or I would show you the once blank piece of paper now covered with a ton of math scribble. If that helps awesome. If not, I'll keep at it until I have an answer for you. Then I'll explain exactly what I did so you can understand.
toastninja  
2 Jul 2009 13:20 | Quote
Joined: 08 Apr 2009
United States
Karma: 2
This sounds like a good physics problem here, its like GS said there will be alot of math involved. On a side note id like to ssee what this contraption looks like if you could post a pic.
JazzMaverick  
2 Jul 2009 19:54 | Quote
Joined: 28 Aug 2008
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uhhhhhggggg.... maths....... I think I'll stick to theory!!
toastninja  
2 Jul 2009 20:29 | Quote
Joined: 08 Apr 2009
United States
Karma: 2
I agree, i never did like physics
blackholesun  
2 Jul 2009 20:33 | Quote
Joined: 04 Jan 2007
United Kingdom
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It's really straight forward. You use conservation of energy. Assuming the system is 100% efficient, the potential energy in the spring = 1/2kx^2 = translational kinetic energy of the plane = 1/2mv^2.

k = spring constant of the spring. To find it, you can hang an object of mass m from the spring, and k = mg/x, where g is 9.81ms^-2, and x is the extension of the spring with the mass on it.

x = extension/compression of the spring. The amount the spring is compressed by before firing.

m = mass of the glider.

v = initial velocity of the glider.

Putting all this together, the initial velocity of the glider is the square root of (kx^2/m)




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