Computational Lab (Short Report)

Motorcycle JumpBrian Bannon (08332321)

November 2012

Abstract

Projectile Motion is by definition,a form of motion where a particle (called a pro-jectile) is thrown obliquely near the earth’s surface, it moves along a curved pathunder the action of gravity.This study has been around for many centuries from theGreeks, through the 1600’s with Galileo, and furthered with Newton in 18th century,till the modern day.In this experiment we used modern numerical methods and computing techniques tomodel the trajectory of a projectile ,in this case a motorcycle and rider,and predictedthe outcome with some constraints on the system.But at the heart of this experimentwe will be using the classic projectile motion equations derived centuries ago.

Introduction

In the previous section we mentioned we are modelling a motorcycle jump, firstly wemust build the system through the constraints and conditions we are given:

”The air resistance on a moving object is given by fr = −cAρvv/2, where v(v) isthe speed and A is the cross section of the moving object,ρ is the density of the airand c is a coefficient on the order of 1 for all other uncounted factors.Assuming thatthe cross section is A = 0.93m2, the maximum taking off speed of the motorcycle is67m/s, the air density is ρ = 1.2kg/m3, the combined mass of the motorcycle andthe person is m = 250kg and the coefficient c is 1.”

The trajectory of a projectile is a parabolic motion that craves out the motionas so;

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Fig(1). This shows the xy map of a projectile in motion with its velocity broken down

into its components through the parabola.

The trajactory of projectile is studied in two directions, the ”x” and ”y” direction.We assume a gravitational force in the y direction and a frictional force proportionalto the square of the velocity and directed against the current direction of motion.From the equations of projectile motion we take the co-ordinate equations:

x = V tcosθ (1)

y = V tsinθ − 1

2gt2 (2)

And from these equations we can derive, if a projectile is launched with an originalvelocity v0 at an angle θ to the horizontal line, the throwing range, max height and

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fly time are respectively

xmax =2v2cosθsinθ

g(3)

2sinθcosθ = sin2θ (4)

xmax =v2sin2θ

g(5)

ymax =v2

gsin2θ (6)

T =2v

gsinθ (7)

We then took the velocity and broke it into its horizontal and vertical components,and took a look at the forces acting on the particle while in motion.

Fig(2). This shows the breakdown of the velocity vector into it’s components,vx = viCosθ

vy = viSinθ, and the forces acting on the projectile.

By using Newtons 2nd law,~F = m~a,this force will effect by changing the accelerationof the projectile. It will, in the x direction ,act to decelerate the projectile reducingit’s range. It will also act on the vertical component of velocity slowing the particleon the both upward and downward part of the motion.

d~x

dt= ~v (8)

d~v

dt=

~f(~x,~v, t)

m(9)

~Fr = −cAρv~v/2 (10)

~Fg = −mg~y (11)

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Thus summing up the force acting on the x and y direction∑~Fx = − ~Frx (12)∑

~Fy = − ~Fry −mg (13)

Where mg is the weight on the acting on the projectile.The drag on the projectile is D = 1

2cAρv2.The drag is proportional to v2, but ~v is a

vector so v2 = vvx and vvy. From these equations we set k = cAρ2m

as these are theconstants throughout the model.Therefore

ax = −kvvx (14)

ay = −kvvy − g (15)

We will see now look the Euler-Cromer method how it is used to derive Acceleration,velocity and position iteration equations. Firstly the Euler method must be intro-duced to solve initial value problems.If

y′ = f(x, y) (16)

y(x0) = y0 (17)

Then

xn+1 = xn + h (18)

yn+1 = yn + hf(xn, yn) (19)

where h is the step size of the iteration, and if this is too large of a number then thevalues will be quite off compared to the real values.If we relate this to the mechanicalmodel of the projectile we get,

dv

dt= a(t), and

dx

dt= v(t), (20)

where a(t) = a(x(t), v(t), t) and we let ∆ t be the time interval between successivetime steps.At time tn = t0 + n∆t we find an = an(xn, vn, tn).With the Euler methodwe have

vn+1 = vn + an∆t (21)

xn+1 = xn + vn∆t (22)

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The Euler Method is an example of a first-order method. The Euler Method is alsoasymmetrical because it advances the solution by a time step ∆t, but uses informa-tion about the derivative only at the beginning of the interval. Moreover the accu-racy of the method is limited and frequency its solutions are unstable.Modificationsof Eqs.(21),(22) yields solutions that are stable for oscillatory systems.Such Euler-Cromer method leads to the following difference equations

vn+1 = vn + an∆t (23)

xn+1 = xn + vn+1∆t (24)

this simple modification conserves energy for oscillatory problems(unlike Euler Methodwhich increases energy of the oscillator with time).

Procedure & Results

Firstly what was needed was the tilting angle of the taking off ramp that can producethe longest range. As we see from the diagram below;

Fig(3).As we can see the angle of tilt on the projectile to obtain the largest range has to

be 450.

The angle which produces the longest range is 450.From here with the initial velocityv0 = 67m/s we use Eqs.(5),(6),(7) to find the max range, max height of the flightand time off the flight

xmax = 457.59m

ymax = 228.79m

T = 9.6587s

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Now broke up the velocity vector into its horizontal and vertical components,

v = 67m/s

vx = 47.35

vy = 47.35

The step size was calculated by dividing the total time by 1000,

N = 1000

T = 9.6587s

∆t =T

10= 0.0096587

Hence, using Eqs(14)(15) computed the new velocities and new positions in the xand y direction for the Euler-Cromer method and wrote the Matlab program(SeeAppendix) to iterate this using the following formulas,

vx(∆t) = v0x + ax∆t

vy(∆t) = v0y + ay∆t

andx(∆t) = x0 + v0x∆t

y(∆t) = y0 + v0y∆t

And the result achieved was

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Fig(4).Graph of the mototcycle jump in the xy plane from Matlab, using the

Euler-Cromer Method.

Lastly was to find what the range of the projectile would be if the landing rampat the finish was 2m above ground?We take Eqs(1),(2)

x = v0xt = 47.37t

y = v0y −1

2gt2 ≈ 1 + v0yt− 4.9t2

y = 1 + 47.376t− 4.9t2

Now you let y = 2, so we get

2 = 1 + 47.376t− 4.9t2

Rearrange4.9t2 − 47, 376t+ 1 = 0 (∗)

Now using the quadratic formula on (∗)

t =−b±

√b2 − 4ac

2a

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we get t1 = 9.647s or t2 = 0.0205s. We ignore t2 because this is when the projectilereaches 2m in the parabola and not 2m above the landing zone, so t = 9, 647 andusing x = 47.37t = (47.37)(9.647) ≈ 457.03m

References

Graphics 1. ”Physics for Engineers & Physicists”-Michael E. Browne (McGRaw-Hill)

Books 1. ”Mechanics for A-level”-L.Bostock, S.Chandler (STP)

2. ”Fundamental Applied Mathematics”- O.Murphy (Folens)

3. ”Computational Physics”- N.J. Giordano (Prentice Hall)

Appendix

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