Yes, the (total weight) = ( car weight ) + ( car gravity weight )...Well, I know now that I will need to get torque for at least twice’s the car weight if I whan’t it to climb up surfaces lol.
I don't understand what you mean by car weight and car gravity weight. It seems you are misunderstanding something...although I am not sure what. In my equation, one term is the force of friction and the other is the force of gravity acting along the slope. THe force of friction is dependent on the normal force, but the force of friction is not the "car weight" as I think you are calling it.
THe whole point of wheels is so you dont have to apply a force equal to the vehicle's weight to get it moving. THat's what wheels do, make it so you can move the thing without lifting it. If it's you need to apply double the weight to roll/move the thing, you might as well pick it up and carry it and reduce the force by half (it could be more only if you are pushing against something, but this analysis is just to get the vehicle to move). To get something moving on wheels, it should always be the weight or less...because otherwise it would require less force to just lift the thing up and carry it...and wheela are meant to reduce the force needed to move someting.
That said, you shouldn't say this you "need twice the torque as the weight" because they are not even the same thing- they have different units. Torque and weight/force are two different things. Torque involves the moment around on a lever or at the cente of a wheel (if you don't understand these words, then let's just say it's torque is "rotational force"). Force is the linear force being applied at the end of the lever (or the edge of the wheel). THat is why after finding the force per wheel you must take into account the size of wheel to convert this force to torque (the larger the wheel, the longer the lever arm and the more torque is required at the center of the wheel to produce the same force at the edge of the wheel). What you should have said was that you needed to apply 2x the weight of the vehicle to get it moving (this statement is still wrong though, read the above paragraph).
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Just by looking at your numbers I can tell something is wrong. Your numbers get really big and then get really small again. YOu should be using 2kg not 2000g, since kg is an SI unit (like N, degrees/radians, and meters). So your units don't match up.
EDIT: OH! I'm sorry. I understand why you did what you did. g in the equation is a VARIABLE for the acceleration of gravity. so Mg = Mxg. It does not mean M is in grams.
Anyways, I get F = 4.92N, Fwheel = 1.23N, T = 9.2mN
Of course, this force is more if you want to push something or accelerate faster. It's just a rough value for the minimum torque that the motor needs to be able to supply to get the thing barely moving. For a given amount of torque, smaller wheels will carry more but move more slowly than larger wheels. Larger wheels, with their larger radiius and corresponding lever arm will move faster but exert less force at the edge of the wheel. So if having big wheels is a bad idea if you have a relatively weak motor. It's the tradeoff between torque and speed. Given that the motor can only exert limited amounts of power, you have to decide where you want this power to go- torque or speed. You can't have both without increasing the power- conservation of energy.
the car gravity weight is proportional to the angle of the inclination that it is trying to go, in that case the ore I try to climb the more the car weight’s, like on the airplanes when pulling G’s , quite interesting, I wish I had a teacher like you for math and physic’s …
Hm. I never though about it that way- defining weight as the normal force rather than the normal force on flat ground (which is usually what is implied).