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Plz help (horizontal spring launcher)


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#18 freeztar

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Posted 31 March 2009 - 07:14 PM

What Modest says above is true.

Nonetheless, I'm going to go out on a limb and suggest that you determine the mass of the catapults' arm. Doing so should give you a value to factor into the Mk. This, along with accurate measurements via database/graphing should get you close enough to find a general use equation. (I think :))

But, as Modest notes, you are introducing much more complexity which means that your results might vary quite considerably.

#19 meleapen

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Posted 31 March 2009 - 08:28 PM

so theres no way that i can calculate how far i have to pull back the arm (so how far the spring will be stretching)?
because the type of spring i got isnt the kind that you compress to get force (ie the kind that have spacing between each... spiral?)
The springs i got are tightly wound, so when you pull back the arm, and you let go, the springs snap back to it's original state, using that force to launch the projectile.

So this is what my catapult looks like. the ________ is the base, a and b are two planks of wood parallel to eachother, perpendicular to the base. the ------------- in the middle is my "axis" made by a rod going through a and b that swivels. so the dots that are kinda diagaonal, is the arm. and the ooo's, represent the spring. So the springs are horizontally coming back from a and b, so when you pull the plank back, they stretch. when you let go, it snaps back and it launches.

hopefullly that makes sense. so with knowing that.. is there any way i can calculate with the spring?


a
| . | b<-this isnt workingbut the"|"should be inline w/ the others.
|ooooo . ooooo |
|--------.----------|
| . | <.... this one should be inline too
|_____________ |

#20 meleapen

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Posted 31 March 2009 - 09:11 PM

found this on yahoo
is this correct?
--



Well, the catapult launches the object at a certain velocity (v), directed 45 degrees upward. For projectiles, you have to break the velocity into an upward and a horizontal velociy. The upward velocity is v sin(45) and the horizontal velocity is v cos(45), so they are both .707v. The reason you do this is because the projectile travels to its highest point with its upward velocity in the same amount of time it travels half of its range with its horizontal velocity. You can find this time (t) using the formula v=vi+at: 0=.707v - 9.8(t), so t =.072v.
In this time, the object travels half of its range, so in twice this time the object travels its full range (50 meters). d = v*t, so 50 = .707v(2)(.072v), 50 = .1018 v^2, so v = 22.2 meters/second.

Now you have the velocity that the catapult must launch the object with. Now the other calculation you have to make is how much you have to compress the springs. This is a conservation of energy problem: when the cataput is released, the potential energy of the compressed springs becomes rotational kinetic energy that sets the catapult in motion, and launches the object.

The equations you use for this are:
Energy of compressed spring: E= 1/2k x^2
Rotational Kinetic energy: KE = 1/2 I w^2 ( I is the "moment of inertia", w is the angular velocity)
and
v=r*w (straight line velocity = angular velocity * radius of rotation)

So, you know that v = 22.2, and r is the length of the arm of your catapult (imagine that the catapult rotates in a circle with that radius), so you can solve for w.

now that you have w, you can find the rotational kinetic energy (1/2 I w^2). I in this case = 1/3 * r * m. r is the length of the arm, and m is the mass ( 1 pound = .45 kg).

Now that you have the rotational kinetic energy, you know what the spring energy has to be.

rotational ke = spring energy = 1/2 k x^2
(if you have more than one spring, your formula becomes rotational ke = n*1/2*k*x^2, where n is the number of springs)

If you don't know k, the spring constant of your springs, then what you do is tie an object (with mass m, in kg) to the end of your spring, hold it in the air, and measure how far the spring stretches because of it (in meters). Now use the formula F=k*x, where F is mg, and x is how much it is stretched), and you can find k.

Now you can solve for x, the distance you need to compress the springs for the catapult to get the right velocity.

#21 modest

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Posted 31 March 2009 - 11:36 PM

I think Freeztar is correct, and Meleapen, wow! Very nice find! :agree:

It's not a general purpose equation—it rather solves a specific case of distance. But, with a little reworking I think we can eek out a single equation with all your needs :D Then again, I'm starting to write this post without yet working through the solution so there's every chance that by the end of this post I'll be saying "I guess that didn't work...". But, let's give it a go.

Ok, the range formula should be more useful than the method the guy on Yahoo used, it is:

[math]R = \frac{v^2 \ sin(2 \theta)}{g}[/math]

We will also need the potential energy of a spring,

[math]E_P = 1/2Kx^2[/math]

And the kinetic energy of your rotating lever (or catapult arm). This will be a tad more tricky, starting with rotating kinetic energy:

[math]E_{RK} = 1/2 I \omega^2[/math]

We need I (moment of inertia) and [math]\omega[/math] (angular velocity). The are given on the links provided. I for the case of a rod rotating about its end is:

[math]I = \frac{1}{3} m L^2[/math]

And angular velocity is,

[math]\omega = \frac{v}{r}[/math]

We substitute I and [math]\omega[/math] into [math]E_{RK}[/math],

[math]E_{RK} = \frac{1}{2} \left( \frac{1}{3} m L^2 \right) \left( \frac{v}{r} \right)^2[/math]

Simplifying this I notice L is the length of the lever as figured for moment of inertia and r is the radius (length of lever) as figured for angular velocity. They appear to cancel giving us:

[math]E_{RK} = \frac{1}{6} m v^2[/math]

which almost seems too simple to be correct, but I can't figure any place that I've gone wrong, so forging ahead... the potential energy in the spring is transferred to the lever when it is released making,

[math]E_P = E_{RK}[/math],

[math]\frac{1}{2}Kx^2 = \frac{1}{6} m v^2[/math]

Solving for velocity squared so that we can plug it into our range equation,

[math]v^2=\frac{3x^2K}{m}[/math]

Plugging [math]v^2[/math] into the range equation given above,

[math]R = \frac{\left( \dfrac{3x^2K}{m} \right) \ sin(2 \theta)}{g}[/math]

solving for x,

[math]x = \sqrt{\frac{Rgm}{3Ksin(2\theta)}}[/math]

Freeztar, you were so freakin' right! It's the same as before except the new mass of the arm and spring constants were factored in. Wow! Very nice call Freezy.

So, ok Meleapen. The private message I sent the other day about finding the spring constant—that's the same. If you have two springs then you will double the spring constant, and if you have three then triple it. The angel theta is the angel between the floor and the line at which the ball leaves the catapult. Also, the distance you pull the arm back should be measured where the springs are attached to the arm. The measurement is actually the extension of the springs. The equation with all the variables listed is:

[math]x = \sqrt{\frac{Rgm}{3Ksin(2\theta)}}[/math]

  • x = distance arm is pulled back in meters (as measured at height of springs)
  • R = Range (distance to target in meters)
  • g = acceleration of gravity (9.8 m/s/s)
  • m = mass of arm and ball in kilograms
  • K = spring constant times number of springs in newtons per meter
  • [math]\theta[/math] = angle of release in degrees or radians depending on calculator used
I'm pretty confident this is correct, but I would appreciate if somebody double checked the math and physics to be sure.

~modest

#22 meleapen

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Posted 01 April 2009 - 03:42 PM

im basically building this; in a larger scale; and where the rubber band is; im going to use a spring

so will the calculations that you guys AWESOME-LY posted work for that type of catapult?


And what does that w symbol mean? it looks like a curvy w (angular velocity?)
what does that represent
and the I's and K's in the energy equations; what do those mean?





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#23 modest

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Posted 01 April 2009 - 04:20 PM

im basically building this; in a larger scale; and where the rubber band is; im going to use a spring

so will the calculations that you guys AWESOME-LY posted work for that type of catapult?


You bet.

That one looks like it shoots at a very shallow angle. In your design I'd make sure you can fire it at a good angle upward.

And what does that w symbol mean? it looks like a curvy w (angular velocity?)
what does that represent
and the I's and K's in the energy equations; what do those mean?

You won't have to worry about [math]\omega[/math] (It is, by the way, pronounced 'omega'. It's a lowercase greek letter). It stands for angular velocity as you guess, but you won't have to put any velocities in the final equation I gave (which is the only one you'll have to use). Also, I, which is moment of inertia. You won't have to worry about. They were both intermediate variables needed to derive the final equation. You can read about them if you want by following the links in my last post.

K is the spring constant. It describes the strength of the spring. This is the only equation you'll need with all the variables explained:

[math]x = \sqrt{\frac{Rgm}{3Ksin(2\theta)}}[/math]

  • x = distance arm is pulled back in meters (as measured at height of springs)
  • R = Range (distance to target in meters)
  • g = acceleration of gravity (9.8 m/s/s)
  • m = mass of arm and ball in kilograms
  • K = spring constant in newtons per meter
  • [math]\theta[/math] = angle of release in degrees or radians depending on calculator used

Assuming your catipult will always be set at the same angle, the only variable that will change from shot to shot will be the range R... how far you want the shot to go. The rest of the variables will always have the same number—once you find the numbers for those variable, they will always go in the equation with those same numbers.

So, as you're building the catapult, you will want to find the answers to the variables I just listed. You'll need a ruler that measures in metric and an accurate scale to measure mass.

~modest

#24 DariusPerez

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Posted 06 December 2017 - 07:14 AM

Wow, I didn't count on so interesting and useful forum page. It'll help me to do my homework, I'm sure. And I hope you solved your issue.
 

 
 

Edited by DariusPerez, 06 December 2017 - 07:15 AM.