The Physics Behind Parking

The Physics Behind Parking

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Background:

After a long day of cruising through town with your buddy, the two of you have grown quite an appetite. You spot a McDonalds at the top of a very steep hill. Unfornately a local biker gang must of had the same idea. Encredibly, the only available parking spot is on the street, uphill of nearly 15 Harleys. There are no other restaurants for 100 miles in all directions. Famished, your friend skillfully manuvers his car to the side of the road. The breaks shudder as the car comes to a stop on the hill. He holds the brake, puts it into first, and shuts off the car.

Problem:

Biker's don't take kindly to people messing with their bikes. Your friend is about to let off of the brake. Being the physics major you are, you have to decide whether it's safe to park or safer to go hungry. Your friends car is pretty old, and the e-brake hasn't worked for years. The car will be held in place solely by the moter. There are two mean looking bikers smoking outside entrance. They are watching you so there probably wouldn't be time to make a run for it.

Known



What You Know:

Your buddy's car is classic, and I'm sure he would want to make more than an educated guess. I've done a little investigating to help you out a bit. I took a torque wrench to the motor, and resists it resists aproximately 46 ft*lbs of torque at the crankshaft. After the compression bleeds down this number is reduced to 38 ft*lbs. The cars rear differential has a 3.73 to 1 gear ratio and a manual transmission with a 3.35 to 1 ratio in first gear. The tires are 28 inches in diameter and the gross weight of the car is approximately 2100 lbs. The hill is often travelled by truckers, and on the way up you noticed a sign that said the hill was at a 26 degree angle with the horizontal.





Summary:

When the car is at rest this means it is in a system of static equilibrium. Gravity is pushing forward on the car, and the tires are pushing back on the car via the reaction force of friction in the motor. The steeper the hill, the greater the force of gravity acting on the car, the greater the reaction force in the motor must have. As stated before, the maximum torque that can be applied to the motor before it rotates is 38 ft*lbs.

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This reaction torque is translated through the transmission and differential where it is applied to the wheel. An equation for the summation of torque can be applied to the wheel. When the torque caused by gravity is greater than the reaction torque caused by the motor, the tires will turn and the car will role down the hill.

Known Values:

Max torque at motor: 36 ft*lbs

Force of gravity = 2100 lbs

Force of gravity (parallel to hill) = 2100 [lbs] sin(O)

Force of gravity (perpindicular to hill) = 2100 [lbs] cos(O)

Radius of tire = 14 in = 1.16 ft

Differential ratio = 3.73:1

Transmission ratio (1st gear) = 3.43:1



Free Bodies

If you were to take away the beer bellied bikers, and the smell of french fries this is what you would have. The free body diagrams should look like this:




The forces acting on this car, are the force of gravity (mgcos(O) & mgsin(O)), normal force, and the reaction force of tires acting against gravity.





The way torque is translated from the motor to the tires is illustrated below:



If a force is applied to two disks of enequal size, the torque is proportianal to increase in circumferance. The picture to the right illustrates this. Gear ratios can be compared to rolling discs. Two rolling discs with an angular velocity of of 3.34[rev/sec] and 1[rev/sec] respectively will travel the same distance if the circumferance of the slower one is 3.34 times that of the faster one. The faster moving disc can be thought of as the input into the gear, and the slower moving disc can be thought of as the output. The torque of the output will be proportianal to the angular velocity of the input..

From Circumference to Torque:

r = 2 * 1 * (pi) r2 = 2 * 3.43 * (pi)

torque = t = 2F C* (pi)

=> (t1) = 2F*3.43*(pi) & (t2) = 2F*1*(pi)

[(t1)/2F * 3.43* (pi)] = [(t2)/2F*3.43*1)

(t1) = (t2)*3.43

generally, tf = ti * (gear ratio or omega)

t1 is directly proportional to t2 by the gear ratio.




There are a total of two gear reductions. One at the transmission (3.34:1) and one at the differential (3.73:1). If we multiply the two together we get the total ratio of 12.5:1 from the moter to the tires. This means for every one time the tires turn, the motor turns 12.5 times.

Using the equation to the right we can calculate the max resistive torque at the wheel caused by the motor.

(torque @ motor)*gear ratio = torque at tires

38*12.5 = 450ft*lbs


Torque at Wheel

If we use what we know from the last page, we can set up a free body diagram at the wheel.



We know when the wheel moves, the torque caused by gravity is greater than the max resistance torque (488 ft*lbs) and it will no longer be in static equilibrium. The equation of torque at static equilibrium at the axle is as follows:

torque [ft*lbs] =

448 [ft*lbs] - 2100[lbs]cos(Ø)*1.16 ft= 0

if we solve for theta we can find the maximum angle of incline before the car roles down the hill.



Solution


Torque @ Wheel:

Let t = torque, (tr) = resistance torque[ft*lbs], r = radius of tire[ft}, F = force of gravity[lbs], n = tire vs motor ratio,

p = theta[degrees]

Sum(tr) = (tr)n - Fsin(Ø)r = 0

=> sin(Ø) = (tr)n/Fr

=> Ø = arcsin((tr)n/Fr)

Plugging in our knowns:

=> Ø = arcsin({36[ft*lbs]*12.5[ft]}/{2100[lbs]*1.16[ft]}) = 11 degrees.

The maximum angle of incline before the car roles down the hill is 11 degrees.

The hill is at an angle 24 degrees. It might be safer to go home hungry. Otherwise you might end up with a pack of P.O.'d bikers chasing after you.

# When first considering this problem, I assumed most of the resistance from the motor would come from the two cylinders in compression. This would complicate the problem considerably, because you would have to determine what the force of air acting on the piston is when it is being compressed. However, Mechanical Engineering Prof. Lin informed me, over time the air in the cylinder will bleed out through the piston rings. Most of the resistance to movement in the motor comes from friction of the rings on the cylinder walls and other internal parts. I didn't have background to find all these force theoretically. To solve this problem I just measured the resistance torque of the motor with a torque wrench.
# It's also interesting to note, initially when I took my torque measurment the torque wrench read 46 ft*lbs. If I let it sit a bit, I would get a reading of 38 ft*lbs. This is consistent with the fact that when you turn the motor over the compression builds up. If you let it sit, the compression will bleed back down. I could imagine how this may fool a person when they are parking on a hill. Initially the car may not move, but as the compression bleeds out, the total resistance is reduced, and the car could start moving.
# On another note, In these calculations I assumed the gears, tires, driveline, and other parts were massless (They didn't have a moment of inertia). It is possible to do this because we are only considering when the vehicle is in static equilibrium. As soon as the vehicle begins to move it becomes a dynamics problem. (Sum of torque = Moment of inertia * angular acceleration)
# Any gear assembly has some power loss due to friction. The calculations for frictional losses are somewhat complicated, but Professor Lin advised me in general, they can be assumed to be 90 to 95% effecient. In the scope this excersize this would probably increase the reaction torque at the wheel. The maximum angle of elevation would increase slightly.
# Finally, The important thing to note it is, the large increase in torque is due to gear reductions. They allow a relativally small resistive force to keep a large mass from moving. It isn't intuitive that a motor that I can turn with my hands should be able to keep a whole car from rolling down a hill.




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