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

Catapult EngineeringPilot 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

Vf = aDt Vf = Vo + aDt

Objects fall toward Earth because of a force

called gravity. Acceleration due to gravity (g) is

9.8 m/s2

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!

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

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.


A few formulas

From rest w/ initial velocity

Vf = aDt Vf = Vo+ aDt

Dd = ½ aDt2 Dd = VoDt + ½ aDt2

Vf = 2aDd Vf = Vo2 + 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 ()

heig

ht (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

•An initial vertical velocity, Voy

v0y

v0x



Launch

Angle ()

heig

ht (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.



Launch

Angle ()

heig

ht (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:



Launch

Angle ()

heig

ht (h

) D

dy

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 ()

heig

ht (h

) D

dy

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.


Thank heavens for that

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