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New Member
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Nov 27, 2013, 04:35 PM
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How to calculate torque needed for a vehicle
Hi,
I am building a rear driven rear mounted engine vehicle. The engine is mounted just behind the rear axle (red box). Driver and seat (yellow), fuel tank, steering etc is located in front of the rear axle.
Total weight of the vehicle including driver is estimated to 500 kg of which 150 kg is behind the rear axle and 350 kg in front of the rear axle. Center of gravity for rear and front compartments are marked.
The idea is when accelerating the vehicle, the front axle to lift from the ground.
Now the question: how do I calculate the torque needed for lifting the front axle from the road?
Thank you!
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Uber Member
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Nov 27, 2013, 07:37 PM
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How do you plan to keep it from flipping over at speed if the front is raised?
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New Member
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Nov 28, 2013, 02:39 AM
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@ high angles the rear part will touch the ground and prevent flip over
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Uber Member
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Nov 28, 2013, 08:36 AM
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Your rear better be only a couple of inches off the ground. At speed, surface loading from air pressure will easily exceed any friction. Watch drag racing videos.
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Expert
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Dec 10, 2013, 05:50 PM
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To determine the torque needed to lift the front end off the ground consider the torques acting around the center of the rear axel. You have the weight of the engine in the back acting in the counter clockwise direction about the axel (a positive torque), and the weight in front acting clockwise (negative toque). The torque of the engine acting through the tires to the ground causes a CCW torque, For the system to be balanced such that the nose of the car is free to lift all three torques must add to zero:
T + 150 Kg x 0.2m x 9.9 m/s^2 - 350 Kg x 1.2m x 9.8m/s^2 = 0,
T = 3822 N-m.
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New Member
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Dec 11, 2013, 02:49 PM
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Dear Ebaines,
Thank you for taking time to respond. I was thinking in the same direction and necessity of quite large engine till I saw a video clip showing an old vehicle with no "positive torque" but even had the old engine mounted in front of the front axel.
In the video you can see when the car moves backwards and then puts in to gear and accelerates forward the front axel lifts that much that the car fillips over backwards.
Please take a look at this you tube link.
MOULOUD_AU_PARIS_DAKAR_(L).wmv - YouTube
How can this possible? I don't think the engin can deliver even 100 Nm.
Regards
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Expert
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Dec 12, 2013, 06:43 AM
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First, you have no idea how powerful that engine is. Second, note that the two men are seated just behind the axle, so there is some positive torque due to their weight. Third, note that the rear of the car falls into a hole just as the driver applies power, thus causing the car to begin to rotate vertically - this gives it some angular momentum to help it start to turn over. Fourth, they're on a hill, which combined with the hole and the compression of the rear suspension all conspire to move the CG of the engine closer horizontally to the rear axle and the CG of the men further behind the axle, thus reducing the amount of torque required to turn it over. And finally note that the car is inbitially moving backwards and the driver applies the brakes to slow it - brakes can apply a lot of torque (the dceleration of a vehicle from brakes is typically much greater than the acceleration you get from the engine), and in this case they help the car to rotate.
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New Member
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Dec 12, 2013, 07:28 AM
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Fair enough.
Once again thanks for your help :)
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