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Trebuchet
Name: Alex G.
Status: student
Age: 13
Location: N/A
Country: N/A
Date: 2/26/2003
Question:
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.
Replies:
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 www.google.com, or go
to one of these pages:
"The physics of the Trebuchet"
http://www.geocities.com/SiliconValley/Park/6461/trebuch.html
"The Algorithmic Beauty of the Trebuchet"
http://www.algobeautytreb.com/
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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