Catapult problem

[Roadam]

Recently I tried to make some kind of a catapult at home. The problem is that when I calculated formulas for it, I found out that it doesnt matter if my boulder is on a long or short arm. But actual tryings made it wrong. I dont know what I did wrong.

 

Construction:

I took a wheel of a bike(small one), put it on an axis and screwed on it a stick about 2r long. Then I took some rope and hang 10kg weight below the wheel. So that when weight lost its potential enegy it was transwered into the energy of a stone I put on the other end of the stick. That is my catapult. First resoults were bad, so I made the stick longer and then it kicked the stone about 10 meters onward.

 

My equations: I suppose that force on the stone is always the same. And the work done onto the stone is force times the arc. A=F*S

F(weight)*r(w)=F(stone)*r(s).

so.. F(s)=Fw*rw/rs

S(arc)=2.35(135° in radians)*rs

when I put those two things toghether r(s) excludes itself!!??

[Tim_Lou]

perhaps you can post a diagram explaining how your device work and how it looks like?

[Roadam]

http://img47.imageshack.us/img47/5031/catapult6zx.th.png' alt='catapult6zx.th.png'>

[Farsight]

Roadam, if you mentally remove the wheel and look at how far the weight drops, what you've got there is a lever.

[Jay-qu]

Catapults are dangerous, to much tension in the wood. Trebuchets are much better :) and they can throw things much further and accuratly ;)

[Qfwfq]

What you reaaly need to reckon on is the velocity the projectile will reach and the angle at which it'll leave the cup, as these are the prime thing in determining trajectory given the value of g (next is air resistance, of course). Without air resistance, and supposing flat terrain, the best angle for distance is 45 degrees.

 

My equations: I suppose that force on the stone is always the same. And the work done onto the stone is force times the arc. A=F*S

F(weight)*r(w)=F(stone)*r(s).

so.. F(s)=Fw*rw/rs

S(arc)=2.35(135° in radians)*rs

when I put those two things toghether r(s) excludes itself!!??

This is true, as you are calculating an energy equal to the work of gravity on the weight! This work won't all go into the stone, indeed a lot of it will become kinetic energy of the weight itself and also of the other moving parts. You'll want to minimize the moment of inertia of wheel and stick and I'd say you need to optimize things for a given mass of projectile.

 

Perhaps this is more of a Science Project though?

[Roadam]

Even trebuchetes have tension in wood. My question in fact is what veilocity does the projectile reaches when thrown. I ceritainly have minded the 30° angle of firing for the best ranges. And parts arent put in much stress.

 

How would you put this into a formula?

[Tim_Lou]

formula for the velocity? sure, simply equation the work done by the mass to the sum of rotational energy of the wheel and the kinetic energy:

[math]{1\over{2}}m{v}^2+{1\over{2}}I\omega^2+{1\over{2}}M{v}^2=Mgh[/math]

 

rearrange:

[math]v=\sqrt{{2M{gh}}\over{m+I/r^2+M}}[/math]

 

edit: almost forgot about the final kinetic energy of the hanging mass, fixed it.

 

it is assumed that there is no change in height of the thing being thrown (you can simply add another term mgh to fix it) and that there is no slipping.

[Tim_Lou]

see, the thing is, in this device the highest velocity the object can reach is:

[math]v=\sqrt{{2{gh}}[/math]

since according to the equation, as M becomes very very big, the limit becomes the above quanity. i would suggest you to pull the string, or use some rubber band or something... a hanging mass? it just has a lot of limitations. you can make h very large, of course...

[cwes99_03]

Ok, sorry to barge in fellas, but can everyone take a look at that device again. This is not a catapult, but a modified trebuchet. The weight, the arm, the rotation. Am I seeing that right?

 

But anyway, yes a trebuchet is just two levers, fulcrum, and two weights.

 

You know how far the weight drops. Calculate the distance "in degrees" from the distance the weight drops that the wheel turns, and divide that by the time the weight takes to fall. Assuming you have a bar to stop the motion of the arm when the weight reaches its minimum potential, thus launching the projectile, you can calculate the tangential speed of the arm at arm's length and have a decent amount of knowledge about how far the projectile will fly based off of the angle of launch.

 

You vary the angle of launch by shortening or lengthening the string attached to the weight, and by adjusting the stop point of the catch bar. I would always place the catch bar where you want the projectile to be fired from then measure out the string and attach it to the weight sitting on the ground. This will help keep you from destroying your device.

 

The further you pull back your arm the greater the time of acceleration and the further you will launch it. The larger the ratio between the launch weight and the projectile the faster the projectile gets to that launch speed.

 

Is that enough info for you?

[Roadam]

In your equations I see that r of the lever(arm) doesnt affect the veilocity. What would then explain my experiments? Or am I wrong? Mybe I just have to add the energy of the lever, so that I would get somekind of a diffrence beetwen r of the wheel and of the lever.

 

And of course this is what I imagined to improve the trebuchet design with. I just didnt know what to call it.

[cwes99_03]

No, R and r are both important (that is of the arm, and of the wheel).

[Roadam]

Well important it is how much they are important and how.

[cwes99_03]

h=r(theda) h=height in meters of the mass drop; r=radius in meters of tire; theda = degree of rotation of tire in radians

R(theda)/t = tangential velocity of the projectile when it leaves the lever arm

R= lever arm in meters; theda =theda from step 1; t=time of acceleration (you can measure this with a stop watch, or calculate it from the time it takes the weight to drop from a height of h, but this is more complicated since it is counterweighted, best to just measure time t)

[Jay-qu]

cwes I do think your right, the first time I go in this thread the image didnt load, but assuming it was a pic of his catapult I suggested a treb, when infact it does work of the priciples of a treb anyway..

[Tim_Lou]

actually, if r=R, then the value of r does not matter.

let's modify the equation assume the moment arms are different:

let 1 be the thing being thrown, 2 be the wheel.

 

[math]{1\over{2}}I_1{\omega}^2+{1\over{2}}I_2\omega^2+{1\over{2}}M{v}^2=Mgh[/math]

 

 

rearrange:

[math]v=\sqrt{{2M{gh}}\over{I_1/r^2+I_2/r^2+M}}*{R\over{r}}[/math]

edit: sry about the error, before the v was the tangential velocity of the wheel, not the thing being thrown. (it was midnight when i post this).

 

since I1 is related to R, the ratio of R to r is involved.

 

well, the change in height of the thing is ignored once again, since it is not known what the initial angle was. it will be quite complicated to generalize it.

[Roadam]

Well, since I planned to use fixed throwing arc(135°) all there is to add probably is change in height of stone and lever times mass and g under the friction-line. I agree that calculating height difference of something that rotates is very complicated, at lest for my level of knowlenge.