Solve Trebuchet Efficiency Mystery

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SUMMARY

The discussion focuses on calculating the energy efficiency of a trebuchet using specific parameters: height (h = 0.71 m), mass (m = 1.445 kg), gravitational acceleration (g = 9.8 m/s²), and time (t = 0.7 s). The theoretical yield of gravitational potential energy (Eg) is calculated to be 10.05 J, while the actual kinetic energy (Ek) is determined to be 43.28 J, resulting in an initial efficiency calculation of 430%. The participant questions the validity of this efficiency and is directed towards a more accurate method for calculating trebuchet efficiency based on actual measured range versus theoretical range.

PREREQUISITES
  • Understanding of gravitational potential energy (Eg = mgh)
  • Knowledge of kinematic equations for projectile motion
  • Familiarity with kinetic energy calculations (Ek = 1/2 mv²)
  • Basic principles of mechanical efficiency
NEXT STEPS
  • Research the formula for trebuchet efficiency: Actual measured range / Theoretical range
  • Study the impact of counterweight mass and drop height on trebuchet performance
  • Explore advanced kinematic equations for projectile motion analysis
  • Investigate methods for measuring actual range and performance of trebuchets
USEFUL FOR

Students in physics, engineering enthusiasts, and anyone interested in the mechanics of trebuchets and energy efficiency calculations.

thua
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Homework Statement


I've calculated the time, the distance, the height of the counterweight before launch, the mass/weight, and I need to calculate the energy efficiency.
How to go about it?

h = 0.71 m
m = 1.445 kg
g = 9.8 m/s^2
t = 0.7 s
x = 4.44 m

Homework Equations



Eg = mgh
Vix = x/t
y = Viy(t) + (1/2) a(t)^2
Ek = (mv^2)/2

The Attempt at a Solution



THEORETICAL YIELD:
h = 0.71 m
m = 1.445 kg
g = 9.8 m/s^2
Eg = mgh
= (1.445 kg)(9.8 m/s2)(0.71 m)
= 10.05 J

ACTUAL YIELD:
t = 0.7 s
x = 4.44 m
Vix = x/t
= (4.44 m)/(0.7 s)
= 6.34 m/s

y = Viy(t) + (1/2) a(t)^2
0.71 m = Viy(0.7 s) + (1/2) (-9.8 m/s2)(0.7 s)^2
Viy = 4.44 m/s

Vi^2 = (Vx)^2 + (Viy)^2
Vi^ = (6.34 m/s)^2 + (4.44 m/s)^2
Vi= 7.74 m/s

Ek = (1/2) mv^2
= (1/2) (1.445 kg)(7.74 m/s)^2
= 43.28 J

43.28 J / 10.05 J = 430%

I don't think my trebuchet is 430% efficient.
What did I do wrong?
 
Last edited:
Physics news on Phys.org
Efficiency is work in/work out
 
An easier way to find trebuchet efficiency (all credit goes to Rom Toms http://www.trebuchet.com/):
Actual measured range in meters /theoretical range= efficiency

theoretical range in meters= 2 * (counterweight in kg * counterweight drop in meters)/ weight of projectile in kg
 
Last edited by a moderator:

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