Catapult Engineering
Pilot Workshop
LA Tech STEP
2007 - 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.
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
V
f= aDt V
f= V
o+ aDt
Objects fall toward Earth because of a force called gravity. Acceleration due to gravity (g) is
9.8 m/s
21 sec 9.8 m/s 2 sec 19.6 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!
0 m/s
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
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
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
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
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
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?
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?
Yes, A curved one. This is PROJECTILE
MOTION.
Look at the next slide for an animation of
these concepts.
A diagram showing the “components” of motion for the projectile launched with horizontal velocity, for
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.
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?
Diagram of a projectile launched at an
upwards angle with an initial velocity of Vo.
A few formulas
From rest w/ initial velocity V
f= aDt V
f= V
o+ aDt
Dd = ½ aDt
2Dd = V
oDt + ½ aDt
2Vf = 2aDd
2+ 2aDd
Speed: V=Dd/Dt
The following acceleration formulas are based
on or can be derived from Galileo's work:
How might this apply to a Catapult?
Projectile Motion
(Motion in 2 Dimensions)
distance (s), time (t)
Launch Angle ()
height (h)
Oh, yeah. An object launched from a catapult is a projectile.
•It is launched with
•an initial velocity, Vo
•An initial horizontal velocity, Vox
v0y
v0x
distance (s), time (t)
Launch Angle ()
height (h)
v0y
v0x
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.
distance (s), time (t)
Launch Angle ()
height (h)
v0y
v0x
Sin = Voy / Vo Voy = Vo Sin
= 22.0 m/s x Sin 40.0o
= 14.1 m/s
Cosin = Vox / Vo Vox = Vo cosin
= 22.0 m/s x cosin 40.0o
= 16.9 m/s
First calculate horizontal and vertical components of Vo:
distance (s), time (t)
Launch
Angle () height (h) Ddy
v0y
v0x
Now let’s calculate Dt
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 a = g = -9.8m/s2
= -[(2) (14.1 m/s)] /-9.8 m/s2
=2.88 s flight time
distance (Ddx), time (t)
Launch
Angle () height (h) Ddy
v0y
v0x
Now we can finally calculate range.
Range = Ddx Vox = Ddx / Dt Ddx = Vox Dt
= (16.9 m/s)(2.88s)
= 48.67 m
= 49 m
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.
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.
