Catapult Engineering PowerPoint

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. 


From rest 

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. 

w/ initial velocity 

Vf = aDt 

M. Nelson / D. McClellan for LA Tech STEP, Jan. 2008 

Vf = Vo + aDt

Objects fall toward Earth because of a force 

called gravity. Acceleration due to gravity (g) is 

9.8 m/s 2 

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 1 st 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 


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 

example a fired bullet. What is Vyo? 


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. 


Speed: V=Dd/Dt 

A few formulas 

The following acceleration formulas are based 

on or can be derived from Galileo's work: 

From rest 

Vf = aDt 

w/ initial velocity 

Vf = Vo+ aDt 

Dd = ½ aDt 2 Dd = VoDt + ½ aDt 2 

Vf = 2aDd 


Vf = Vo 2 + 2aDd

How might this apply to a Catapult? 


height (h) 

Projectile Motion 

(Motion in 2 Dimensions) 

v 0y 

Launch 

Angle () 

v 0x 

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 


•An initial vertical velocity, Voy

height (h) 

v 0y 

Launch 

Angle () 

v 0x 


A projectile is launched with an initial velocity of 22.0 m/s at an 

angle of 40.0 o . 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. 


height (h) 

v 0y 

Launch 

Angle () 

v 0x 


First calculate horizontal and vertical components of Vo: 

Sin = Voy / Vo 

Voy = Vo Sin 

= 22.0 m/s x Sin 40.0 o 

= 14.1 m/s 

Cosin = Vox / Vo 

Vox = Vo cosin 

= 22.0 m/s x cosin 40.0 o 

= 16.9 m/s 


height (h) Ddy 

Launch 

Angle () 

v 0x 

v 0y 


Now let’s calculate Dt 

Ddy =VoyDt + ½ aDt 2 

= 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/s 2 

= -[(2) (14.1 m/s)] /-9.8 m/s 2 

=2.88 s flight time 


height (h) Ddy 

Launch 

Angle () 

v 0x 

v 0y 

distance (Ddx), time (t) 

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. 



Thank heavens for that

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