# Simple Calculations for Information on a Car

by mjgarrin
Tags: calculations, information, simple
 P: 6 I just bought a new car, nothing incredibly special. And being a nerd, I love creating theoretical calculations based on given data. So, I've been trying to predict the effects on my car under certain scenarios. I am doing my best to figure out the effects based of equations alone, and not taking my car in for a dyno test. Sadly, I am having difficulty with some equations I really hope someone out there can help me. These are the things I would like to determine: 1) If my car is traveling on a flat road at a certain speed, what is the force of the car? 2) If my car is traveling on a flat road at a certain speed, what is the torque that is being generated by the axle shaft? 3) If my car is traveling on an uphill road at a specific angle at a certain speed, what is the force of the car? 4) If my car is traveling on an uphill road at a specific angle at a certain speed, what is the torque that is being generated by the axle shaft? 5) How much torque is being generated by the transmission at a certain engine RPM value? So far, this is what I have come up with, assuming the engine speed is 3000 RPM and I am only in 1st gear: Vehicle Speed (MPH) = (Engine Speed (RPM) x Tire Radius (in)) / (168 x Axle Ratio x Trans Ratio) Vehicle Speed (MPH) = 21 MPH Transmission Speed into Axle (RPM) = Engine Speed (RPM) / Trans Ratio Transmission Speed into Axle (RPM) = 669 RPM Wheel Speed (RPM) = (Vehicle Speed (MPH) x 5280 x 12) / (60 x PI x Wheel Diameter (in)) Wheel Speed (RPM) = 242 RPM Axle Shaft Torque to Wheels (lb-ft) = Trans Torque into Axle (lb-ft) x Axle Ratio Axle Shaft Torque to Wheels (lb-ft) = ????????????????????? Trans Torque into Axle (lb-ft) = Engine Torque into Trans (lb-ft) x Trans Ratio Trans Torque into Axle (lb-ft) = ????????????????????? (I know the Max Torque at a given RPM value, but I want to find the torque at any RPM value) Torque of Axle Shaft (lb-ft) = radias of shaft (ft) x Force (lb) Torque of Axle Shaft (lb-ft) = 0.125 (ft) x ???????????????? Torque of Wheel (lb-ft) = radias of wheel (ft) x Force (lb) Torque of Wheel (lb-ft) = 1.209 (ft) x ???????????????? From a free body diagram, I have determined the summation of forces in the X and Y direction: FY = N - mgcos($\vartheta$) = 0 Fx = F(car) - F(friction) - mgsign($\vartheta$) = ma N = mgcos($\vartheta$) F(friction) = $\mu$N F(car) = ????????????? m = weight of car These are a few calculations I have determined and a few I am having a problem determining theoretically. I would like to find out these values for any engine RPM value and any given speed. Here are some specs of my car: Max Torque: 250 lb-ft @ 4500 RPM Max Horsepower: 265 HP @ 6250 RPM Weight: 4078 lb Axle Ratio: 2.77 Trans Ratio: 1st: 4.484 2nd: 2.872 3rd: 1.842 4th: 1.414 5th: 1.000 6th: 0.742 Rev: 2.882 Wheel Diameter: 29.02 in Vehicle Performace: 0-60 mph in 7.4 seconds I'd really appreciate any help anyone can offer. Thanks!! Mike
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P: 22,313
 Quote by mjgarrin 1) If my car is traveling on a flat road at a certain speed, what is the force of the car?
The question isn't worded correctly so doesn't have any meaning as written. Do you mean the force between the car and the ground? ....which is also equal to the force of wind resistance? Not easy to calculate from scratch. You at least need to measure your fuel consumption rate.
 2) If my car is traveling on a flat road at a certain speed, what is the torque that is being generated by the axle shaft?
Same as above.
 3) If my car is traveling on an uphill road at a specific angle at a certain speed, what is the force of the car?
Same as above, but at least the potential energy part is equal. The power due to the "rise" of the hill is mgh/t. Power is fd/t. You should be able to take it from there, using geometry.
 4) If my car is traveling on an uphill road at a specific angle at a certain speed, what is the torque that is being generated by the axle shaft?
Same as above.
 5) How much torque is being generated by the transmission at a certain engine RPM value?
Torque does not depend on RPM alone.
P: 6
 Quote by russ_watters The question isn't worded correctly so doesn't have any meaning as written. Do you mean the force between the car and the ground? ....which is also equal to the force of wind resistance? Not easy to calculate from scratch. You at least need to measure your fuel consumption rate. Same as above. Same as above, but at least the potential energy part is equal. The power due to the "rise" of the hill is mgh/t. Power is fd/t. You should be able to take it from there, using geometry. Same as above. Torque does not depend on RPM alone.
Maybe I can ask another question then:
How would you calculate the force between the tire and the ground? I know T = r x F, so if I know the radius of the tire, how do I calculate the torque and/or force between the tires and the ground?

Could I then use the torque generated by the tire and determine the torque of the axle shaft?

 Engineering Sci Advisor HW Helper Thanks P: 7,279 Simple Calculations for Information on a Car You can measure the wind resistance, if you can find a level road with a lack of other traffic, and calm weather so any "real" wind doesn't mess up the experiment. Put the car out of gear and measure the rate of deceleration (which will depend on the speed of course). All you need is the speedometer and a watch. The handbook will give the mass of the car, so you can calculate the force = mass x acceleration.
 P: 572 I think this web page might answer a lot of your questions. Look for the theory at the bottom of the page.
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P: 22,313
 Quote by mjgarrin Maybe I can ask another question then: How would you calculate the force between the tire and the ground? I know T = r x F, so if I know the radius of the tire, how do I calculate the torque and/or force between the tires and the ground?
By knowing the wind resistance or the output of the engine.
 Could I then use the torque generated by the tire and determine the torque of the axle shaft?
Absolutely. But finding the torque at the tire isn't any easier! You still have the same problem!

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