Catapult Engineering
Pilot Workshop
LA Tech STEP
2007 - 2008
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Some Background Info
• Galileo Galilei (1564-1642) did
experiments regarding Acceleration.
• He realized that the change in velocity of
balls rolling down inclined planes and
falling objects were accelerated by the
same phenomenon and followed the same
mathematical rules.
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Galileo observed that the final
velocity of an object starting
from rest and accelerating at a
constant rate equals the
product of the acceleration and
the elapsed time. If it had an
initial velocity, the final velocity
will equal the sum of the initial
velocity and the increase in
velocity caused by the
acceleration.
From rest
w/ initial velocity
Vf = aDt
Vf = Vo + aDt
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Objects fall toward Earth because of a force
called gravity. Acceleration due to gravity (g) is
9.8 m/s2
0 m/s
1 sec 9.8 m/s
2 sec 19.6 m/s
10 sec 98 m/s
If a bowling ball
dropped from the roof,
after the 1st second, it
would be traveling 9.8
m/s.
A second later its
velocity is 19.6 m/s.
After falling for 10
seconds, its velocity is
98 m/s or about 219
miles per hour!
That’s fast!
Isaac Newton (1642-1727)
•Newton pondered Galileo’s work and motion
in general
•He realized that the force (gravity) that
caused the acceleration noted by Galileo was
the same force that kept the planets in their
orbits
•Newton formulated three laws of motion
• Two are important right now for us
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
What exactly is a force?
• A force is a push or a pull
• A force can act though contact
– Spring, rope, chain, friction, etc
• A force can act a distance
– Gravity, magnetism, electrical
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Newton’s First Law
• The Law of Inertia
An object at rest will remain
at rest, or an object in motion
will remain in motion with
constant velocity when the
net force acting on the object
is zero
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Newton’s Second Law
The Law of Acceleration
• The effect of an applied force is to
accelerate a body in the direction of the
force.
•The acceleration
is proportional to
the applied force
and the mass of
the object.
F=ma
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
What does this have to do with a catapult?
Hang on, I’m getting there.
• Consider,
– A bullet is fired horizontally from a rifle
– A second bullet is dropped from the rifle’s height
at the exact instant the bullet leaves the rifle’s
barrel.
•Which bullet
strikes the ground
first? Justify your
answer.
Ignore Air resistance
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
What force is acting on the bullet flying
horizontally? Which of Newton’s Laws
applies to this bullet?
What force is acting on the bullet that was
dropped? Which law applies to this one?
How many “components of motion are
applied to the dropped bullet?
How many “components of motion are
applied to the fired bullet?
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
That is right. They strike the ground at the same
time. Why? Because the only force being applied
to each of them in the vertical direction was
gravity. Therefore they fell to the ground at the
same rate.
However, their flight paths (trajectories) are
different.
The bullet that was dropped had a path that
was straight down. What kind of path did the
other one follow?
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Yes, A curved one. This
is PROJECTILE
MOTION.
Look at the next slide
for an animation of
these concepts.
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
A diagram showing the “components” of motion for
the projectile launched with horizontal velocity, for
example a fired bullet. What is Vyo?
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Note, For Projectile Motion:
• In these illustrations there was an independence
of horizontal and vertical motions.
– Horizontal motion is under Newton’s first law;
therefore, it is at constant horizontal velocity
– Vertical motion is under Newton’s second law;
therefore, it is at constant downwards
acceleration
• The combination of these two motions results in
the observed parabolic path of a projectile.
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Now, lets launch the projectile at an
upward angle.
• Again, What forces act vertically? Horizontally?
• As a result, what type of flight path is taken?
• What components of velocity are involved?
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Diagram of a projectile launched at an
upwards angle with an initial velocity of Vo.
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
A few formulas
Speed: V=Dd/Dt
The following acceleration formulas are based
on or can be derived from Galileo's work:
From rest
w/ initial velocity
Vf = aDt
Vf = Vo+ aDt
Dd = ½ aDt2
Dd = VoDt + ½ aDt2
Vf = 2aDd
Vf = Vo2 + 2aDd
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
How might this apply to a Catapult?
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Projectile Motion
v0y
height (h)
(Motion in 2 Dimensions)
Launch
Angle ()
v0x
distance (s), time (t)
Oh, yeah. An object launched from a
catapult is a projectile.
•It is launched with
•an initial velocity, Vo
•An initial horizontal velocity, Vox
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
•An initial vertical velocity, Voy
height (h)
v0y
Launch
Angle ()
v0x
distance (s), time (t)
A projectile is launched with an initial velocity of 22.0 m/s at an
angle of 40.0o. Calculate the range of the projectile.
To calculate range, you need to use this formula: Ddx = VxDt
Therefore, we need to calculate Dt, Vx
But, to calculate Dt, we need to calculate Vy
So let’s get at it.
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
height (h)
v0y
Launch
Angle ()
v0x
distance (s), time (t)
First calculate horizontal and vertical components of Vo:
Sin  = Voy / Vo
Cosin  = Vox / Vo
Voy = Vo Sin 
Vox = Vo cosin 
= 22.0 m/s x Sin
= 14.1 m/s
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
40.0o
= 22.0 m/s x cosin 40.0o
= 16.9 m/s
height (h) Ddy
v0y
Launch
Angle ()
v0x
Now let’s calculate Dt
distance (s), time (t)
Ddy =VoyDt + ½ aDt2
= Dt (Voy + ½ aDt)
since projectile goes up and back down Ddy = 0
0 = Dt (Voy + ½ aDt)
0 = (Voy + ½ aDt) & 0 = Dt
Dt =- [(2)(Voy)] / g
= -[(2) (14.1 m/s)] /-9.8 m/s2
=2.88 s flight time
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
a = g = -9.8m/s2
height (h) Ddy
v0y
Launch
Angle ()
Now we can finally
calculate range.
distance (Ddx), time (t)
v0x
Range = Ddx
Vox = Ddx / Dt
Ddx = Vox Dt
= (16.9 m/s)(2.88s)
= 48.67 m
= 49 m
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Wait
How to you get the projectile up to it’s
initial velocity, Vo?
Right, a force has to be applied to
accelerate the projectile.
That is where the spring comes in.
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
What’s Next?
In order to design and build a catapult to
accomplish certain tasks, you are going to have
to apply kinematic (motion) formulas and solve
for the variables concerning projectile motion,
angular acceleration, potential energy of springs,
and other such stuff..
Fortunately for me, that is someone else’s job to
show you.
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008
Thank heavens for that
M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008