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# BombDiggity101

Best trebuchet ever
by

## louisa kishton

on 2 October 2014

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#### Transcript of BombDiggity101

d=-0.067
(-1.28s+149.7)(-1.28s+149.7)
+5.7(-1.28s+149.7)+13.85
Therefore
sling length
distance
d=distance
v=velocity
g=gravity
=release angle
Distance Equation
Alrighty,
Lets build a trebuchet!
But how?
How will we make it hit our target?
Change Counterweight ?
move the entire trebuchet?
But sling length only changes release angle? How will that change distance?
counterweight

+ arm
+ rotational power
+ ball
+ Friction
potential energy arm
Potential Energy Counterweight
Friction
potential energy Ball
potential energy Arm
kinetic energy Arm
kinetic energy ball
Before launch
After launch
Potential Energy
Kinetic Energy
= too many variables
Finding Potential energy
mass x height x gravity
1. FRAME/base
(32.174 ft/s )
finding kinetic energy
1/2 x mass x (velocity)
2
2
(8.19 kg)(32.174 ft/s)(2.75 ft)+(40.82 kg)(32.174 Ft/s)(2.39 ft)+ W(F) =

(m3)(32.174 ft/s)(10.52 ft)+(8.19 KG)(32.174 ft/s)(7.25 ft)+(1/2)(8.19 kg)(v2^2)+(1/2)(m3)(v3^2)
724.64 + 3091.21 +W(f) = 338.5 (m3) + 1910.4 + 4.1 (v2)^2 + .5 (m3) (v3)^2
1905.45 + W(F) = 338.5 (m3) +4.1(v2)^2 +.5(m3)(vs)^2
1905.45 + W(F) = 338.5 (m3) +4.1(v2)^2 +.5(m3)(vs)^2
plug in mass of a baseball (0.142 kg) into M3
1905.45 + w(f) = (338.5)(0.142) + (4.1)(v2)^2 +.5(0.12)(v3)^2
1905.45 + w(f) = 48.067 +4.1(v2)^2 + .06(v3)^2
1905.45 + w(f) = 48.067 +4.1(v2
2. Axle
1857.38 + w(f) = 4.1(v2)^2 + .06(v3)^2
1857.38 + w(f) = 4.1(
v2
) + .06(
v3
)
but
v2
and
v3
can be solved for with the same variable
3. Crossbar
4. arm
velocity= (constant(
x
5. Counterweight
v2
= (
x
)(2.5)
v3
= (
x
)(8)
6. Sling and trough
plug in these values
2
2
Thanks mr. Chaffee!
1857.38 + w(f) = 4.1(2.5
x
) + .06(8
x
)
but since this is a theoretical value we pretend friction doesn't exist
2
2
1857.38 = 29.465(
x
)
2
63=
x
2
x
=7.9
so velocity of baseball =
8
x
= 8 (7.9) = 63.5 ft/s
2
now we take this value for the velocity
63.5 ft/s
and plug it into the distance equation
2
d=-0.067a +5.7a+13.85
R =0.998
2
2
sling length to release angle
Angle of release
movement of trebuchet
different release
tilt of arm
faulty setup
a=-1.28s+149.7
r =0.826
2
some more boring math
So now that we've proven that
And also That
We Can Finally correlate
Or, In Mathematical Terms
Because
d
=-0.067a +5.7a+13.85
2
And
a =
-1.28s+149.7
Then we can say,
d
=-0.067(
-1.28s+149.7
) +5.7(
-1.28s+149.7
)+13.85
2
But that's pretty complex
simplification powers activate!
d=-0.067(-1.28s+149.7) +5.7(-1.28s+149.7)+13.85
2
d=-0.067
(1.6384S-383.232+22,410.09)
+5.7(-1.28s+149.7)+13.85
d=
-0.1097728s +25.676544s-1501.47603
+5.7(-1.285+149.7)+13.85
2
d=-0.1097728s -118.380544S-634.33603
2
2
d=-0.1097728s +25.676544s-1501.47603
-7.296s+853.29
+13.85
Ahh... That's better
2
WE have to account for
friction
(theoretical-experimental) ÷ theoretical
=
% Distance lost to friction
This works out to be about
0.13
, or
13%
distance lost to friction.
For accurate distances, we should multiply our equation by
0.87
to account for this.
d=-0.1097728s -118.380544S-634.33603

• 0.87
d=-0.0955023s -102.9910737S-551.872346
this is our final equation
HISTORY
y =release height
0
2
2
GOAL
MATH

Correlations
So to sum it all up...
Because we can relate
sling length
to
distance,
we can accurately hit a target of any distance just by
changing the length of the sling.
A BUNCH OF BORING MATH
More
remember this is theoretical Data
Full transcript