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Name: Alex G.
Status: student
Age: 13
Location: N/A
Country: N/A
Date: 2/26/2003

Dear scientist,

I am a seventh grader doing my E-project or science project. I have been bamboozled with trebuchets since I was a little kid. I have been to England to learn about them but did not really mind anything interesting. So I have come to the conclusion that I should ask a scientist. I really want to know how creating the counter-weight closer to the fulcrum increases its velocity? Also if you extend the arm to a longer length is it able to hurdle the projectile further? If you have any formulas for the velocity or anything else please send it to me and explain it briefly. Thank you for taking the time to answer my questions and for your time.

There are several very good web pages on this device, with drawings, formulas, simulations, etc. I could never approach in a this text-only message. Type "trebuchet" into the search box at, or go to one of these pages:

"The physics of the Trebuchet"

"The Algorithmic Beauty of the Trebuchet"

Tim Mooney

A trebuchet is just a giant lever. You have probably used a lever to help you lift something really heavy.

The way a lever works is that it has a short side and a long side. If you are lifting something using a lever, you push down on the long side a little, and the heavy thing goes up. Also, the long side moves much farther than the short side.

So, a lever works basically by having one side move a lot and requiring just a little force, and the other side moving just a little, but providing a lot of force.

A trebuchet is just a lever, but does this in reverse: the short side has a huge weight which drops down, and the long side then moves really quickly, but can't lift as much.

The formula for figuring this out is the length of one side divided by the other, or as a formula:

R = L1/L2

For example, if you have the short side of your trebuchet 1 inch, and the long side 10 inches, and you hang a 10 pound weight on the short side, then you have the following:

10 lb times 1" divided by 10" = 1 lb. And you know that 10 lbs on one side will be balanced by 1 lb on the other side. Or as a formula:

W1 * R = W2

This does not account for the weight of the lever itself, but it gives you the idea, and works well as long as your weights on each side are quite a bit more than the weight of the lever.

Also, with this trebuchet, you can find see that when the end of the 1" side drops one inch, the long end rises 10 inches, so the short end moves 1/10 (0.1), as fast as the long side.

So with a trebuchet, you know that, moving the fulcrum closer to the weight allows you to through things faster (and so farther), but means that you have to throw smaller things.

Also, to figure out how fast it will throw things, you can account for the weight on the two sides using this rough formula:

Weight on short side times ratio minus Weight on long side all divided by Weight on short side times ratio plus Weight on long side times acceleration of gravity. Or as a formula:

(W1 * R - W2) / (W1 * R + W2) * 32.2 feet/(s*s) = Acceleration of short side.

So, in another example: You make a trebuchet with a 1" short side, and a 10" long side, and hang a 10 pound weight on the short side, and use it to throw a 1/10 pound projectile:

R = 1"/10" or 0.1

(10 lb * 0.1 - 0.1 lb) / (10 lb * 0.1 + 0.1 lb) * 32.2f/(s*s) (1 lb - 0.1 lb) / (1 lb + 0.1 lb) * 32.2ft/(s*s) 0.9 lb / 1.1lb * 32.2ft/(s*s)

Which is about 26.3 ft/(s*s). So the short side will fall at about 80% the acceleration rate of gravity while the long side will rise at 800% the acceleration rate of gravity.

I hope this helps you understand how a trebuchet works.

Eric Tolman
Computer Scientist

Bamboozled by trebuchets... I never thought I would hear that from a 13 year old. Good for you, however there are quite a few equations that help explain multiplication of force, velocity, and linear motion. I recommend that you look at your local library for an "elementary" physics book that, at minimum, has a chapter on levers (moment arms), a chapter on angular/circular motion, and a chapter on linear motion. The chapter on levers will help explain why the length of the trebuchet provides a multiplication of force. The chapter on angular/circular motion will help explain the velocity of the object being hurled, the effect of centripetal forces, and the effect of speed on the distance form the rotation axis. The chapter on linear motion will help you calculate where your object will land. One of my favorite physics books (but a little long on explanation) is "Physics Principles & Applications" by Harris, Hemmerling, and Mallmann, but I am sure you local library has a plethora of physics books available for investigation.

Good luck.

Chris Murphy

I am not sure I can answer your question very helpfully, but some basic principles might help. I assume a trebuchet consists of a log rod with a projectile on one end and a weight on the other. Let us call R the ratio of the distance from the weight to the fulcrum to the distance from the projectile to the fulcrum and S the ratio of the mass of the weight to the mass of the projectile. Let us imagine the weight of the rod is negligible.

If RS is much greater than one, the weight will fall almost unimpeded by the projectile weight and so will fall with an acceleration close to g = 32 feet per second squared.. Note that the acceleration of the projectile is then 1/R times g and so decreases as R increases. This explains why having the weight closer to the fulcrum can increase the speed of the projectile. However, RS must remain much greater than one. If the projectile is 10 times further from the fulcrum as the weight ( R = 0.1), the weight must be much larger than 10 times the weight of the projectile for maximum effectiveness (making it still heavier does not help, since its acceleration can never be greater than g.

If R is 0.1 and the mass of the weight is much more than ten times the mass of the projectile, and if the projectile and if the projectile is accelerated throughout a travel distance of 10 feet, the maximum range of the projectile would be 200 feet.

I hope this helps. If you have further questions, I would be delighted to study the problem more carefully. I have also found trebuchet's fascinating.

Best, Dick Plano, Professor of Physics emeritus, Rutgers University

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