Hill Climbing Primer

[SPYDERLK]

You said that by saying the pinion would continue to turn with infinite traction and a resisting load at ground level. Applied at ground level the unmoved resisting load exactly offsets pinion torque while it is holding back the tractor motion. So how can the pinion keep turning?

Now we're getting somewhere.


Your statement above that I hilighted is the issue. It is definitely possible for the pinion to continue turning because we have one motion unaccounted for. The lifting of the front of the tractor. And that is exactly what would happen. If all elements of this demonstration continue, the tractor will do a wheelie. It has no choice.

You can demonstrate this with any tractor. You can raise the front tires of your tractor off the ground and hold them in the air, lift them higher, lower them back to the ground, all by turning the pinion. I cannot count the number of times I've done just that when trying to move an object that resisted my attempts.

Now if you add another element, such as CalG's suggestion that you weld the axle shafts to the housing, or if you securely anchor the front of the tractor, then all motion stops.

In our discussion CalG and yourself never touched on the fact that the front of the tractor is still free to do whatever the applied torque requires. In this case it would be to lift. Put the tractor in reverse and the opposite thing happens, the front would be forced into the ground.

Keep all this in mind, no traction loss, continuous turning of the pinion, no load movement. The inevitable result is the front of the tractor rises. Doesn't take any graphs, diagrams, charts, formulas or any other solutions to prove that.

.......
Lets put some numbers to it so we can check whats happening; 1] Diameter of Ring Gear = 1 foot, 2] Diameter of Drive Wheels = 4 feet, 3] Torque on Axle = 20,000 ft-lb.

,,,,In order to impart that torque to the axle the ring gear face must be pushed downward by the pinion with a force of 40K#. There is the potential for this force, but since we have not applied a load the tractor just glides forward in lo gear. No force since theres no opposition to movement. ... Now we stop and apply a resistance to tractor motion. We apply this resistance to the end of a rigidly fixed lever extending straight down to the ground from the tractor body directly below the axle. The distance from the end of the lever up to the axle is 2 feet - the wheel radius. The tether runs straight back, horizontal and at ground level.

,,,,Now as the tractor moves incrementally the force and torque builds until 20K ft-lb is being imparted to the ring gear ... and consequently the axle and onward to the wheel. Since the wheel has 2' radius the rearward force it delivers to the ground is 10,000#. The tractor tries to move forward, but the tether resists that force with 10K# pulling back. The lever it acts on is 2 feet long so the 10K# force on it produces a torque reacting into the tractor body of 20,000 ft-lb around the axle. This torque acting at the point of pinion/ring gear drive - 6" from the axle center - puts a downforce on the pinion of 40,000 lb.
,,,,,,,,,,,,,,ANY other set of numbers would produce the same balanced result. Its the way it works.
larry

 

[Egon]

.......
Lets put some numbers to it so we can check whats happening; 1] Diameter of Ring Gear = 1 foot, 2] Diameter of Drive Wheels = 4 feet, 3] Torque on Axle = 20,000 ft-lb.

,,,,In order to impart that torque to the axle the ring gear face must be pushed downward by the pinion with a force of 40K#. There is the potential for this force, but since we have not applied a load the tractor just glides forward in lo gear. No force since theres no opposition to movement. ... Now we stop and apply a resistance to tractor motion. We apply this resistance to the end of a rigidly fixed lever extending straight down to the ground from the tractor body directly below the axle. The distance from the end of the lever up to the axle is 2 feet - the wheel radius. The tether runs straight back, horizontal and at ground level.

,,,,Now as the tractor moves incrementally the force and torque builds until 20K ft-lb is being imparted to the ring gear ... and consequently the axle and onward to the wheel. Since the wheel has 2' radius the rearward force it delivers to the ground is 10,000#. The tractor tries to move forward, but the tether resists that force with 10K# pulling back. The lever it acts on is 2 feet long so the 10K# force on it produces a torque reacting into the tractor body of 20,000 ft-lb around the axle. This torque acting at the point of pinion/ring gear drive - 6" from the axle center - puts a downforce on the pinion of 40,000 lb.
,,,,,,,,,,,,,,ANY other set of numbers would produce the same balanced result. Its the way it works.
larry

Sounds good but what allows the pinion gear to exert force?

 

[SPYDERLK]

There has to be drive power on one side and a force resisting motion on the other .Force relies on there being a counterforce.

 

[Egon]

There has to be drive power on one side and a force resisting motion on the other .Force relies on there being a counterforce.

So what happens when the drive power is greater than the resisting force?

 

[SPYDERLK]

So what happens when the drive power is greater than the resisting force?

If the power links to the ground the load moves. The force applied is only as much as needed to overcome the resistance.

 

[Egon]

If the power links to the ground the load moves. The force applied is only as much as needed to overcome the resistance.

Fixed rpm, gear drive and the pinion turns at a constant speed. So what happens if the forward movement is not as fast as the pinion is delivering? Tires may slip, they may not? What if they don't?

 

[SPYDERLK]

.......
Lets put some numbers to it so we can check whats happening; 1] Diameter of Ring Gear = 1 foot, 2] Diameter of Drive Wheels = 4 feet, 3] Torque on Axle = 20,000 ft-lb.

,,,,In order to impart that torque to the axle the ring gear face must be pushed downward by the pinion with a force of 40K#. There is the potential for this force, but since we have not applied a load the tractor just glides forward in lo gear. No force since theres no opposition to movement. ... Now we stop and apply a resistance to tractor motion. We apply this resistance to the end of a rigidly fixed lever extending straight down to the ground from the tractor body directly below the axle. The distance from the end of the lever up to the axle is 2 feet - the wheel radius. The tether runs straight back, horizontal and at ground level.

,,,,Now as the tractor moves incrementally the force and torque builds until 20K ft-lb is being imparted to the ring gear ... and consequently the axle and onward to the wheel. Since the wheel has 2' radius the rearward force it delivers to the ground is 10,000#. The tractor tries to move forward, but the tether resists that force with 10K# pulling back. The lever it acts on is 2 feet long so the 10K# force on it produces a torque reacting into the tractor body of 20,000 ft-lb forward tip around the axle. This torque acting at the point of pinion/ring gear drive - 6" from the axle center - puts a downforce on the pinion of 40,000 lb.
,,,,,,,,,,,,,,ANY other set of numbers would produce the same balanced result. Its the way it works.
larry

Fixed rpm, gear drive and the pinion turns at a constant speed. So what happens if the forward movement is not as fast as the pinion is delivering? Tires may slip, they may not? What if they don't?

IF hooked at ground level [or below] AND the tires didnt slip, the pinion would stop OR twist off the axle. The tractor would not tip back at all.

,,,,,,,,,,,,,, I want to add: It is very important that no one try the exact scenario described in the numerical example above. With the pull point extending straight down from axle center the balancing forward torque is exact but the situation is critical and un damped. In the unlikely case that traction and drive forces are real life huge, a NON steady state pull - from quick acceleration like popping clutch - could start a backtip thereby tipping the pull lever forward; in effect raising it. Now the pullpoint is above ground and there is net backtip. ... I set the example up with the lever at the axle position for clarity. For safety that same lever should be located a few inches behind the axle so it would descend if any backtip occured.
larry

 

[Egon]

IF hooked at ground level [or below] AND the tires didnt slip, the pinion would stop OR twist off the axle. The tractor would not tip back at all.

,,,,,,,,,,,,,, I want to add: It is very important that no one try the exact scenario described in the numerical example above. With the pull point extending straight down from axle center the balancing forward torque is exact but the situation is critical and un damped. In the unlikely case that traction and drive forces are real life huge, a NON steady state pull - from quick acceleration like popping clutch - could start a backtip thereby tipping the pull lever forward; in effect raising it. Now the pullpoint is above ground and there is net backtip. ... I set the example up with the lever at the axle position for clarity. For safety that same lever should be located a few inches behind the axle so it would descend if any backtip occured.
larry

Me thinks with you're levers at equilibrium and a torque applied at the pinion that is equal or greater than the tractor center mass to centreline of rear axle the front will continue to rise. You got a few more lever's there in the dynamics of things that don't diss appear in some gear case housing!:thumbsup:

 

[SPYDERLK]

Me thinks with you're levers at equilibrium and a torque applied at the pinion that is equal or greater than the tractor center mass to centreline of rear axle the front will continue to rise. You got a few more lever's there in the dynamics of things that don't diss appear in some gear case housing!:thumbsup:

,,,,,,,,,,,,,, No.

 

[Egon]

,,,,,,,,,,,,,, No.

Yes , it does happen that way.