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Front to Rear Weight Transfer; some simple demonstrations

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  • Front to Rear Weight Transfer; some simple demonstrations

    This is a simple demonstration that you can do yourself. You will need an analog scale, or one that will measure changing weight, a Load, and a rope used to pickup the Load.

    Your body will represent the different parts of a car. Your feet are the stationary rear tires, your body and legs are the wheels, your shoulders are the rear axle bearings, your arms are the chassis, your muscles are the force, and the Load is the front tires, plus the weight of the car that is on the front tires.

    Figure A;

    Standing on the scale with your arms outstretched at 90° to your body, the Load is sitting on the floor, and the rope is slack. The weight on the scale is your body plus the rope, the weight on the rear tires of the car.

    Figure B;

    Raise your arms slowly and lift the Load a half inch off the floor and stop. The weight on the scale now includes your body, the rope, and the Load, the weight of the whole car on the rear tires.

    Figure C;

    Rapidly raise the Load until your arms are at a 45° angle. The weight on the scale is the weight of the whole car plus the weight created by the force of lifting the Load.

    The added weight created by the force of lifting the Load, is only present at the rear tires, on the scale, while the Load is being lifted.

    The amount of added weight is determined by the weight of the Load and chassis, the distance the Load is from the rear axle center line, and the force used to lift the Load.




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    For this demonstration a picture of a slot car chassis with the wheel base of a Slot it 312 is used. The motor makes 200 g/cm Ts with a 2.91:1 gear ratio.

    The instant the power is applied to the motor, it produces 667 grams of force at the 6mm pitch circle of the red pinion.

    This 667 grams of force is transferred to the 17.5mm pitch circle of the red spur gear. This creates a force of 584 grams at the 20mm tires to accelerate the car forward.



    The instant the power is applied to the motor, it produces 667 grams of force at the pitch circle of both red gears. The pinion gear attempts to climbs the stationary spur gear, this creates a lifting force at the front wheels of 83 grams. The chassis becomes a third class lever.




    When the rear tires start to rotate, the lifting force begins to diminish. (like trying to run up the down escalator)

    Part of the force is trying to rotate the rear tires forward, and part of the force is trying to lift the front of the car. Whichever gives the least resistance will accelerate the greatest.

    When the motor starts to rotate, the motor’s force begins to diminish, until it becomes zero at maximum rpm.

    The 6mm pitch circle and 17.5mm pitch circle are the imaginary non slipping mating surfaces of all of the 6.5mm and 18mm gears.


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    In this video demonstration, the rear tires of the car are attached to a base and cannot turn. The front wheels are sitting on a Magnet Marshal. A 200 gram weight is placed on top of the car, and the scale is zeroed.

    When 4 volts is applied to the motor, the motor creates a lifting force of 26 grams, or 78 grams @ 12 volts, at the front wheels. The motor in this car made over 180 g/cm Ts, with a 3.09:1 gear ratio. The motor heats up rapidly and the force starts to drop immediately.

    [ame="http://s164.photobucket.com/albums/u20/davejr-photos/?action=view&current=MYSTERIOUS3.flv"]http://s164.photobucket.com/albums/u20/davejr-photos/?action=view&current=MYSTERIOUS3.flv[/ame]

    Since the wheels and motor can’t turn in this demonstration, only the forces created at stall can be measured.

    An inline motor and gears create the same forces, only the motor and pinion gear are turned 90° to the chassis, and the teeth on the spur are at 90°.

    Dave
    Last edited by davejr; 02-19-2009, 07:12 AM.

  • davejr
    replied
    Brockfan

    I believe figure C in the stick figure drawing, rapidly raising the Load, covers all of your questions.

    To take it further, once the Load is elevated to 45° and comes to a stop, the weight on the scale returns to just the weight of the car. But when the lifting force is decreased, and the Load starts to be lowered, the weight on the scale decreases below the actual weight of the car.

    With a normal slot car, the amount of front wheel lift is governed by the pickup braids. If the are adjusted properly, the motor power is reduced enough to control the wheel stand and keep the pickup in the slot.

    Dave
    Last edited by davejr; 02-19-2009, 02:34 PM.

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  • brockfan
    replied
    Dave, won't the scale show a higher reading while you are actually lifting the load....and the faster you lift the load, the higher the spike will be as you will also be overcoming increased inertia?....as tied to hp being a given amount of work done in a given time....
    In a slot car, this inertia would be a very high spike in the first millisecond, after that, as motor torque drops off, so does the rate of acceleration and so does the inertia.....infact, inertia may actually become an counteracting force to motor torque lifting the front of the car as the rate of acceleration decreases....
    ????

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