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Name: Mike
Status: Student
Grade: 6-8
Location: WI
Country: United States
Date: October 2006

I have been interested in medieval times since I was 7 years old. My dad, brother, and I are building a trebuchet, it so far is 13 feet high and is not yet anchored into the ground. Now before we try to launch something I want to know a little about how much should be in the counter weight in equivalent to the load. We are planning on launching things that weigh about 20-30 lbs. For an example, if I were to launch 10 pound objects, how much weight should be on the other end?

The answer to your question is very dependent upon the engineering details of your weapon. For the good of your device, and the good will of your neighbors, I think you should start with small weights and counterweights and empirically determine the load / counterweight / range. The key is to start with small light projectiles!

Vince Calder

Mike- fun project, but be real careful. The heavy ballast could collapse the machine and throw parts nearby at any time. Hopefully you can arrange to cock it with a rope, from a little distance. And being behind the machine in line with its throw is a danger zone during cocking and throws.

I do not know from personal experience or formulas, but when I have seen trebuchets on TV the ballast-to-projectile ratio always seemed to be large, 10:1 and up. That means your 20 lb., uh, watermelon, would need a 200 lb ballast, minimum. Going bigger, say 400lb, could get to be a big strain on the frame. You might find it more impressive to shoot a 10 lb melon with the same 200lb ballast.

Use a "conservation of energy" principle to give you an idea.

Suppose the trebuchet was tuned to fling the load straight up. (It cannot really shoot that direction efficiently, but just imagine so.) If the whole arm is 13 feet long, the ballast swings down maybe 5 feet. Energy to lift an object is E = M x H, Mass times Height. E = (200 lb) * (5 ft) = 1000 ft-lb. (Foot-pounds are a respectable old English unit of work or energy.) If your machine has a perfect swing, exactly all that energy goes into lifting the load into the air, so 1000 = E = (20 lb) * (?? ft.) ; ??= 50 feet. Science predicts it cannot go higher than 50 feet. It will usually do some amount less.

Then imagine turning your perfect vertical throw to 45-degrees diagonal, the angle that gets maximum distance. That means half the energy is directed upwards, and half sideways. The vertical half gets the melon to 25 feet high. The horizontal half of the energy just keeps the projectile travelling until it is stopped by hitting the ground, so it will go something like 100 feet downrange. (For a given launch angle there is a constant ratio of distance-to-height.)

The 10-lb object could go twice as high and far on the same energy, but you may need to re-tune the machine's swing for every weight-change, to make that happen right.

It is also possible that a trebuchet is just easier to tune up to ideal energy-transfer efficiency for lighter loads or a load-to-ballast ratio higher than 10.

Jim Swenson

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