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Trebuchet CATAPULT Project/Worklog Updated 12/12/06 Please COMMENT

Discussion in 'General' started by scarecrow, 21 Apr 2006.

  1. Stuey

    Stuey You will be defenestrated!

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    Exactly! What you want is for the top of the arm to move as far as possible during the tie the weights fall. Making the arm longer will make the arm more further in the same amount of time. Greater distance over the same amount of time = greater velocity. Greater velocity = greater distance.

    Don't forget that you want to launch at an angle of 45° for greatest range, but you can tune that part once everything is built and stable.

    Torque = radius *cross* force. For your purpose, *cross* can mean *times*. If you were to have the beam horizontal, and placed the beam such that the counterweights were placed 1 foot away from the pivot, 100lbs would balance out a 10 lb weight10 feet away. For a 14 foot beam, if your counterweights are let's say 2 feet away from the pivot (this is reasonable I believe), then you'll ned 120 lbs to counteract the 10 lbs at the opposite end (12 feet from pivot).

    I'm not sure what ratio is optimal for trebuchets, but I think I read a 4:1 ratio is optimal. If that's the case, then you'll need only 500 pounds to launch your projectile. If you want even greater range, then add more weight or add more weight and increase the length of the arm.

    I'm a bit too bust to calculate the exact estimated speed with which the projectile would be launched with, but assuming the weights fall 1/2 meter (it'll actully be more but I don't feel like whipping out converters and such), and an acceleration due to gravity of 10 m/s (rounded up).

    Distance = 1/2 a t^2 (simplified since initial velocity is 0). So 0.5m 5 m/s^2 t^2.

    0.5 / 5 = t^2, so t = 0.316 seconds

    Now... let's say our beam starts off tilted so that the weights are positioned 30° above the horizontal. The tip of the throwing arm will be 30° below the horizontal. Let's say that's exactly where the projectile is. If it launches at 45°, then it will have travelled 75°.

    Change in angle over change in time is equal to the angular velocity. Angular velocity us equal to v/r. So we have 75° = 1.31 radians. So... 1.31 rad / 0.316s = v/r. With r being 10 feet (approx 3 meters), velocity of the tip will be 12.4 m/s. I *think* I did this all correctly.

    Of course this is with a massless arm and it doesn't take into account the projectile weight or the counterweight weights. The acceleration isn't really 9.8m/s^2 here (or 10 as I rounded it) since the projectile will be counteracting it. So instead of 9.8m/s^2, it will be less. The calculations become messy b/c torque has to be involved, but I'll get into that in a few days if nobody else does.

    The range as such would be about 50 feet, I believe. That estimate seems far too low. Can anyways point out where/what I did wrong? The range should be v^2 sin(2theta)/g, which would be v^2 sin (90) / g. It should be about 15 meters and change, or roughly 50 feet.

    Hmm... of course the projectile will be attached via a sling (right?) so that will also greatly increase the velocity as it increases the radius between he projectile and the pivot.

    If you add another 1 meter of rope extending to a sling, you'll get an extra 30+ or so feet out of the range. Additionally, you're goingto use more than 0.5m for the counterweight length, so that should increase the acceleration time.

    Before you do anything else, sit down and crunch these numbers. I've given you most of the equations you need. What you have to do is figure out the range you need (300 feet?) and the beam lengths you need for it. I figure your acceleration will not be 9.8 but maybe 8.0 or so after compensating? Since you're doing this for physics, calculate the torques and by how much it takes away from the downward force of the counterweight, and that should give you your true acceleration.


    EDIT: I just saw the last posts after I posted this so I didn't see the corrected beam values. I'm too lazy to recaluclate everything, but these values should still offer a *rough* estimate.
     
  2. scarecrow

    scarecrow What's a Dremel?

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    yeah you are way to low you got to put more weight on. the cw to projectile ratio is 100 to 1. So 100 pounds for a 1 pound projectile. That is not taking into account the mass of the thowing arm. The numbers will be exactly crunched after my ap physics exam. Lol in the meantime I am just looking at logistics of supporting this. According to rough calculations I did a while back the range will be 240 feet. I will crunch the numbers when I get some time. For now I got a butload of other physics work to do. I will post my calculations here when I do them latter for all intrested.
     
  3. Stuey

    Stuey You will be defenestrated!

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    Nonono, I didn't mean counterweight to projectile ratio of 4 to 1. I meant counterweight to *effective weight* ratio. In other words, 500 pounds to a 10 pound projectile if the projectile is at 12 feet and the counterweight at 1 foot. So the ratio I assumed was about 40 to 1. Perhaps it is still a bit low.
     
  4. scarecrow

    scarecrow What's a Dremel?

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    Well I will do calculations latter and post them up here. I think something is flawed in your formula's anyway just don't have the time currently to find it. But I did do some rough calculations back in janurary that said the range would be around 240 feet. I will do it again just right now AP's crazyness so I am just gathering parts for the axel and structure.
     
  5. Stuey

    Stuey You will be defenestrated!

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    I miscalculated the angles by a LOT. Instead of a 75° rotation, it should be approx. 120° or so! :wallbash:

    Although I know that the projectile is launched at a 45° angle, I had assumed that the projectile launches when its holder was on the beam. I had not taken into accout that a sling trails the arm by a little bit and that the arm is near verticle when the projectile is launched. Thus the range is a lot more than I had calculated. BUT all is not lost as at least the equations are now available to those who want em and don't want to google for em.
     
  6. scarecrow

    scarecrow What's a Dremel?

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    thanks a lot man very usefull. I wonder does anybody have any ideas for a name. Or anything else I should do with this project?
     
  7. MaximumShow

    MaximumShow Minimodder

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    THIS is what you want. And HERE is a free one. Check it out.
     
    Last edited: 24 Apr 2006
  8. scarecrow

    scarecrow What's a Dremel?

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    dude the free one is for an ongar and I know the other but I can just do the math myself its not that bad. Just as I said need to do it after the tests cause I am too busy studying.
     
  9. Constructacon

    Constructacon Constructing since 1978

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    This is great. Final pics/video are a MUST.

    For the target car you could approach a local auto wrecker. If this is getting fired at a sporting event or getting coverage in your local paper they may be willing to sponsor you with a free car. Alternatively you could place an ad in your local paper asking for a free car. You don't care what it is and some people do have junkers they want to get rid of (My parents gave away some rusting heap to anyone who would take it off their hands). MAKE SURE YOU REMOVE THE GLASS BEFORE YOU SHOOT AT IT.




    On a side note, I've been thinking about these a lot recently because a beer company has a MASSIVE one of these in their ads. Watch the video. Makes me laugh every time I watch it.
     
  10. scarecrow

    scarecrow What's a Dremel?

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    Nice constructacon that is hilarious. I wish it rained beer would be the best thing ever. So you guys get how big my trebuchet is going to be. Total weight of machine around 2000 pounds. Height just shy of 20 feet not in firing position and 8.5 feet when in firing position. Length 17 feet about. Width about 9 feet. :) So yeah big. I am going to be taking pictures and explaining every step of the way. So if you plan on making your own you can learn from my mistakes and successes. I need to clear shooting at targets for now its just going to be watermelons and we want to see how far we can launch a football with this thing :). Football might be too light but if it does work can you say Field Goal from 200 yards off. So yeah this will be the first worklog I know of for a treb enjoy guys and please feel free to comment.
     
  11. DXR_13KE

    DXR_13KE BananaModder

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    i remember in AGE OF EMPIRES there was 2 famous trebuchets, something like "god's sling" and "bad neighbour", and instead of dremeling the name, that would cause a weak point, why dont you vinil it.
     
  12. scarecrow

    scarecrow What's a Dremel?

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    Ok got an idea for the axel. Since I don't have access to car bearings or anything of that sort I am going to be using the bench press bar. But to reduce friction I was thinking of putting another pipe around the bench press. The pipe will have an inside diameter of 1 inch. So it will fit snug but not tight around the bench press bar. Then grease up the inside. So it will be metal on metal instead of metal on wood. Plus the grease or axel oil will work better and the wood won't soak it up. What you guys think of the idea and if you have a better one please say. I been looking for actual bearings though no luck :(. So yeah going to have to make/improvise something.
     
  13. George_Buchanan

    George_Buchanan What's a Dremel?

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    I hope it's strong enough - it wouldn't be very nice if it self-destructed with anyone/thing in the vicinity.

    is it just me or do lots of people know a freakishly large amount about these things
     
  14. Lovah

    Lovah Apple and Canon fanboy

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    If you are getting a car to shoot at.. you could just steal a axle from that donar car to build your trebuchet. The less friction the bearing gives, the better the trebuchet will be.
     
  15. scarecrow

    scarecrow What's a Dremel?

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    Small update got some equipment to use for the construction. The list if first from me 1 handdrill, my dremel for cutting bolts to size and axel to size YAY DREMEL BEST TOOL EVER, 2 hammers and 2 ratchets with a set behind it not going to list all the sizes not necessary. Also going to bring a bunch of drill bits for holes don't know if one fits yet don't have the bolts yet. Lets see obtained from other sources. 1 circular saw. 1 belt sander and 2 more handdrills. And 1 nice handsaw.

    You might be wondering why all the handdrills well as I said a lot of my wood is going to be wood used already and still attached to other stuff. Put together primarily with screws so more people unscrewing faster I get my hands on my wood. So yeah just a tiny update and I said construction starts after AP's which is May 8 so start 9th. If everything goes well.
     
  16. Stuey

    Stuey You will be defenestrated!

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    While I have limited experience with a dremel, I have even less experience with car axles. But, what I do know is that I think you're going to go mad trying to cut through it with a dremel. You *may* want to consider finding an angle grinder with a metal cutoff tool, but I have no fact to back that up with; it's just a random opinion. It's basically like using dremel cutoff wheel, but one that's about 10 times larger, more durable, and a bit thicker so less chance of shattering. If you have a shop/tech teacher in your school, you may want to consult them.
     
  17. scarecrow

    scarecrow What's a Dremel?

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    I took your advice stuey I found one actually I was going to resort to dremel last resort. Hurray for my school artroom they have one so I am going to use that instead. Thanks for the advice man ;).
     
  18. Stuey

    Stuey You will be defenestrated!

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    No problem! I just visualized that it would be like cutting an oak tree down with a butter knife!
     
  19. TheoGeo

    TheoGeo What are these goddamn animals?!

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    You should add wheels! The wheels will absorb alot of the force once the projectile has been launched and stop the whole thing shaking itself apart. The wheels shouldnt move much untill the projectile is launched due to the forces being balanced untill then and therefore wont decrease force behind it.

    There was a program where they built a trebuchet a while ago on tv and they discovered that adding wheels somehow helped with range slightly too. I think its something to do with it stabablising the whole thing and stopping it rocking around
     
  20. scarecrow

    scarecrow What's a Dremel?

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    Yeah it was a nova program I saw it. The swinging counter-weight is much more effective though. And less complicated. So I am doing that instead.
     

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