Any rate.
Wrong.
a (traction) pull coefficient of 5 is the minimum needed, the faster you want to accelerate the higher that will go.
Wrong.
F = ma. How many times do i have to write that formula down for you? Since F = ma, it follows that a = F/m
If I have a 100 kg mass I can accelerate it with any force whatsoever. If I apply 1 Newton force I can accelerate it at 1/100 m/s^2. If I apply 10 N of force I can accelerate it at 1/10 m/s^2. If I shove it with 1,000,000 Newtons it will accelerate at 10,000 m/s^2. There is NO 'minimum force' required to accelerate any given mass, ignoring all other factors such as friction, rolling resistance etc. The thing that determines minimum force is any force that acts directly to oppose the motion the wheel is trying to impart. Since the wheel is trying to move the vehicle forward, the limiting force has to be one that acts 'backwards'. Weight does not do that on the level surface. Weight acts straight down. There is NO lateral component of weight that opposes forward motion. Therefore, weight is NOT directly a limiting factor in accelerating a vehicle forwards at any rate. It does of course affect things like traction and static friction, but you certainly do not need to exert a force equivalent to its weight just to get the vehicle moving. Hence I can push a car.
Please note that when I am talking about pull coefficient or drawbar pull I am talking about the formula in the army test, so it is not a measurement of force, it is a measurement of the mass the wheel can pull before traction is useless.
No, it really is not. That is YOUR erroneous conclusion. I have already explained, twice, what those numbers actually represent and how that test applies to reality.
Drawbar pull, anywhere, is the force available to pull a load after the vehicle has moved itself. That's what it is in all cases, INLCUDING this test. You can keep on saying it means something else here, but the authors of that paper would not agree with you.
The wheel has a 57lb load on it, on the moon it needs enough traction to pull 342lbs to accelerate, or 6 times the wheel load, which equates to a drawbar pull of 5.
No it does not.
Once again for clarity, in the army test the pull coefficient is the amount of mass the wheel can pull (above and beyond it's own load) before traction becomes useless, it is not a force.
Which bit of what I have written about what those numbers actually represent is not clear to you?
So yes, the wheel does not have to be able to pull with a force anywhere near its own weight to move, but it does have to be able to get enough traction to pull 6 times its weight.
No it does not. It ONLY has to pull against its own weight when the rover goes up a slope and its weight, which is an acceleration itself, is acting against the force the wheel is applying to move the rover forward.
No the results show the wheel has enough reserve traction to go up a 25 degree slope on earth only.
No, they don't.
When the testing shows a pull coefficient of 0.5 before traction is too problematic, it is saying the wheel has enough traction to drive the weight on it plus 50% more (which is enough to get it up a 25 degree slope here on earth).
But on the moon it needs enough traction to drive/pull the weight on it, plus 500% more.
No, it does not.
Do you see what i am saying?
Yes, but that does not make you any more right.
A wheel on earth with 57lbs of weight on it has to have enough traction to drive 1/4 of a vehicle weighing 228lbs.
On the moon a wheel with 57lbs of weight on it has to have enough traction to drive 1/4 of a vehicle weighing 1368lbs.
And how much traction is required to drive that?
Stop trying to redefine already well-defined parameters to make your argument fit your conclusion. drawbar pull is a measure of force, FULL STOP. That force does not have to exceed the weight of the vehicle to move it along the surface, ever. That's why I can push a car, a tractor can tow an airliner and a train can tow several carriages.
