Hill Climbing Primer

[My Hoe]

Only on page 18 (and admittedly, I skipped the first 8 or 9) but stunned enough by what I'm reading that I must subscribe. FWIW, I'm with ovrszd.

 

[Egon]

Good. Absolutely, level ground for the simple case. [We can handle slopes too, but why complicate matters when we are stuck on the very principles of Mechanics.] Now pick out the stuf you need to know to simulate it identically. I know its hard - all those twirling gears and bearings. But when you understand it its easy. For an easy example see Post#291 and 297 where there is a small edit for clarity.
larry

Using yourself for reference?

 

[Egon]

Re: Hill Climbing Primer
<Back Flips:

Enough torque, enough traction and the tractor will rotate around the rear axle. Don't mater how low you hitch.
Egon
50 years behind the times
Livin in a
Worn out skin bag filled with rattlin bones >

My original statement on backflips. Simple and to the point. No unrealistic conditions.
No flex shafts or welded ring gear.

If someone wishes to disagree that is their privilege.

 

[SPYDERLK]

Using yourself for reference?

Cant use the ones who are wrong. It would be counterproductive. When are you going to assess the situation accurately so that you can get just the pertinent parameters into a simulation?
larry

 

[ovrszd]

.......
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

Very well stated. Only problem is it has NOTHING to do with my statement concerning the front of the tractor becoming light......

 

[ovrszd]

There is no doubt the front end will lift some, absolutely. Right around 45 degrees I think you would approach some steady state where the forces are going to balance. The stated conditions really don't exist because no tractor has infinite torque or traction. The load could certainly appear so. But even if the conditions did exist, it couldn't go all the way over because to do so would require forward motion of the hitch, requiring motion of the load and thereby breaking one of the initial conditions.

I'll ask again: can God make a rock so heavy he can't lift it? There is no answer to this question because it is a contrived situation.

I agree in your scenario the tractor cannot tip over backwards.

Good to see someone that actually understands and accepts that the front of the tractor tries to lift.

You have put yourself in the opposing court of CalG and Larry.

 

[ovrszd]

Only on page 18 (and admittedly, I skipped the first 8 or 9) but stunned enough by what I'm reading that I must subscribe. FWIW, I'm with ovrszd.

You will find yourself in hazardous waters where there's much circling and gnashing of teeth by the opposition!!!

 

[ovrszd]

Cant use the ones who are wrong. It would be counterproductive. When are you going to assess the situation accurately so that you can get just the pertinent parameters into a simulation?
larry

There ya go again...... Calling the opposition "wrong". I don't get that. Again I say, I can only assume that means you have no further evidence to prove your point so you accuse the opposition of error??

I'm fairly sure Egon and/or myself have never called you "wrong" or any of the other condescending terms you've used in reference to us. But then I'm sure if we did you will copy/paste every one of them in your reply, along with several of your previous statements as if we might have missed reading them.

 

[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 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

Very well stated. Only problem is it has NOTHING to do with my statement concerning the front of the tractor becoming light......

It addressed that implicitly rather than explicitly. - - Below I have made a few descriptive additions to help bridge the concepts to see the actual result:
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#. [so, the ground, in resisting this force is exerting 10K# to the 2' wheel radius -- causing a rear tip torque of 20,000 ft-lb on the tractor] 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 also, so the 10K# [backward] force on it produces a forward tip 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. The tip forces cancel and the tractor pulls neutrally with no weight transfer.
,,,,,,,,,,,,,,ANY other set of numbers would produce the same balanced result. Its the way it works for a pull point at ground level in a steady state pull -- no acceleration. ... For a pull point above ground level the anti - backtip torque would be proportionately less as described below:
larry

The "dueling levers" take everything into account. To duel with the thrust lever, the load lever [resisting applied thrust- aka pull point] must extend below the axle. Both levers, thrust and load, act at the axle. When the load lever is short a hi proportion of torque resisting tip comes from front weight bias. As the load lever is lengthened more and more resistance to tip is produced [a forward torque] and massive front weight becomes less important in real world.-- Nevertheless, in our posited infinite traction and output torque circumstance the front weight would not prevent tip.-- This situation continues until the load lever is lengthened enuf that it places the pull point at ground. ... Now, as usual, all force from the thrust lever is reacted equally against a force on the load lever, but now of the same length, pulling back opposite to thrust. The torques, whether small large or infinite, are always equal and opposite. There wouldnt be a twitch. No change in F/B weight on the wheels. Nothing.
... If you think there is a force unaccounted for it may help to see that the thrust lever [tire radius axle to ground] is just a direct extension of the ring gear.

larry

 

[CalG]

Open letter to ovszd

I just went through these many posts to be sure of your position. I wanted to confirm that my "oposition" was directed accurately. I do not want any of my comments to be felt to question any person, only the technical truths. To be clear, I oppose the idea of "the pinion climbing the ring gear" (Ref your post #59) only on the grounds that it is an indirect description and not applicable to all situations. Cases such as tractor designs without ring and pinion of any type (vis. hydraulic hub drive, etc..) The analogy of an internal mechanism somehow "lifting" the external member that contains it is just not my idea of complete.
You confirmation of the "No ouptut condition" in your posts # 109 and #153 suggests that you also hold that the pinion forces do not actually do any lifting, but rather only transfer forces to some external member.

You acknowledgement of a "high enough" hitch point on your post # 98 shows that you are aware of differing reactions to load and pull point. But then you reverse your position in post #233 and posit that a tractor will wheelie regardless of hitch point. This I disagree with, even though I know that in practice it can often appear that way. In real situations, one seldom pulls from below ground level ;-) In your most recent posts , such as #315 you have softened this unconditional position to a more moderate "front of the tractor becomeing light". Perhaps it is only a matter of degree for you? But even this is not satisfactory to me, for with some hitch points, forces of sufficient magnitude can be generated that would press the front end more firmly downward. Adding to the front if you will.

Your dismissal of my offers of extreme conditions (such as welding the axle) demonstrate a position based on limited observation and example.
I happen to grasp concepts more readily when the extreme condition is explored. I recognize not everyone visualizes as I do ;-)

In your post # 280 you stated two contentions. Though answering your question seemed an unproductive challenge to me, an answer was offered regarding the forces on the pinion bearing, and is still waiting your response. It will need to be seen as to where that might go.

Alas, now in post #316 you have moved away from your previous position on irresistable forces and the enevitable flip over, and now agree that "the tractor can not tip over backwards".

What a change.!

Now, I am only "an imaginary person on the internet" (your post #152) but I regard these technical conversations as valuable if there is open communication, and something can be learned.
Your continued openess and direct sincerity are appreciated!

This post is quite a bit off topic. My desire is that by recounting my observations of this discussion, there will be some understanding gained. If I have erred in my interpretation of the referenced statements and posts, please correct me.

On the technical side of this topic, I have done some additional doodling. It is easy enough to draw lines and arcs to show that unless all the load lines between the load fixed point and the tractor pull point are in a straight line, there will be a "lifting force" on the tractor front end in some attitudes, and a restoring force in others. The generation of restoring forces are responsible for opposing the torque reactions that would otherwise tip a tractor over backwards. We are most familiar with the ones that result in lifting because that's the way we most commonly hitch to loads ;-). The two wheeled tractor users must have developed a high level of sensitivty for these qualities.

kind regards

Cal