This question isn’t directly related to the game this year but for other use. It can however be applied, just be aware the game is not the intent of the question.
I am just curious as how to determine torque that is needed to be applied to wheel(s) to overcome rolling resistance. For example, My motorcycle only has 16 foot-pounds of torque, yet…me and the bike manage weigh 500+ pounds and hit 80+ mph. I know the transmission has alot to do with the max speed (obviously) but I am looking for concepts/equations to help determine the force needed to overcome rolling resistance. Maybe to apply to something such as how much torque is needed to propel a 200 pound object to 30 mph.
I think it would be useful for you to review the various forces and terms and come to grips with how they relate to each other. I am not an engineer, so take this with a grain of salt. Ultimately, it takes a certain amount of energy to propel an object at a certain speed through some medium (air?) with some system losses due to friction and whatnot. How you get there can vary. You may have barely enough torque to get things moving and accelerate, but plenty of power to move fast eventually ( a 50 cc. road racing motorcycle). Why do the land speed record vehicles take forever to accelerate to top speed? Why do big diesel trucks shift through 3 gears to get to 15 MPH, or so it seems?
Anyway, do dig around. If you can find some advice on what kind of power you need to overcome friction and stuff to propel a bike at 30 MPH, that is good. I never weighed 200 pounds and it’s been awhile since I’ve propelled myself at 30 MPH.
Let’s look at this from an energy perspective. Energy is conserved, so in your robot/vehicle/whatever, energy in = energy out. Your motors are producing a certain amount of power, that’s your energy in. So of that energy must be used to overcome various forces that resist motion. For example:
air resistance. as your robot moves, it smacks into air molecules and imparts some energy to them. typically, the air resistance increases as you go faster, usually force is proportional to velocity squared. Example: at about 20 mph in your car, air resistance is pretty low and your radio, air conditioner, and headlights are probably using more power than air resistance. At 60 mph (3 times faster) air resistance is 9 times higher, and dominates all other power consumption in your vehicle.
friction, viscous drag, and internal damping inside your transmission and drive train. None of the components in your drive train are 100% effecient and all of them suck a little bit of power. This power ends up becoming heat. These kinds of forces also usually go up as velocity goes up.
internal damping in your wheels (aka hysteresis) for a car, this can be significant. as the wheels roll, the part of the tire near to the ground compresses a little bit. so the tire is not quite perfectly round anymore, it’s got a tiny little bit of a flat spot on it. As the wheel rolls, that part returns to round and another little bit is deformed. It take some energy to do this, and again that energy ends up as heat. For the robot this year, the wheels are pretty stiff, so that probably won’t be a bit factor. Again, these forces will be higher as velocity increases.
there are other forces involved, but these are the big ones.
Now, let’s think through a couple of different cases.
case 1) velocity = 0, motor power = 100W. since all of the above forces are zero, all of the available motor power goes into increases the vehicles kinetic energy. the vehicle will accelerate quickly
case 2) velocity = 50%, motor power is stil 100W. But now, since there is some velocity, you have all of the above forces which suck up some available energy. hypothetically, let’s case 75W are consumed this way. 25W is still available to accelerate your vehicle, so it continues to accelerate, but slower.
case 3) velocity = 100%. motor power is still 100W. but now, the drag forces are consuming 100W. there is no power left to increase the kinetic energy so the vehicle no longer accelerates.
I hope that gets you started. there’s lot of good articles on wikipedia and places like that if you want to dig into it…
I am aware of the theory. I am looking for something or someone to explain it in terms of physics and math. The theory is simple I am aware that all transmissions has drag etc. I have taken a full year of physics in college and 2 full years of math in college. I am thoroughly equipped, I just would rather not derive it myself since it’s been done already. I have looked around the internet already but to no avail, perhaps I am referring to it by the incorrect name.
I am unsure how to be sure everybody knows what I am talking about. If you look on wikipedia for rolling resistance it speaks of wheel/contact surface deformation. I am not speaking of this. I am look for a simple way to mathematically model, (or measure) the max speed of a body with n number of wheels and a torque T being applied to m wheels. Or something similar if this exactly isn’t so easy.
It’s pretty hard to come up with an equation that will tell you what force it will take to overcome rolling resistance and aerodynamic drag…generally these numbers are found empirically, and then an equation is made up with the necessary constants.
Here are some equations I swiped from the internets, it’s a NASA page about hybrids so I think it’s public domain…
A-1. Aerodynamic Drag Losses
The following equation calculates power loss due to aerodynamic drag, represented by the variable Paero(Watts).
Eq. for aerodynamic drag P_drag = (A_frontal x C_drag x (Density) x V^2)/2
Afrontal = Vehicle frontal area (m2)
CD = coefficient of drag of vehicle
V = vehicle velocity (m/s)
rhoair = atmospheric density, kg/m3
The following equation calculates loss required to overcome rolling resistance, represented by Proll.
Keep in mind that torque is only one part of the motor-power equation. If you’re talking about torque, maximum RPM is important as well in order to determine an approximation of how much horsepower you have. Since motorcycle engines typically have very high redlines, it means that so long as you keep them revved quite high, it’ll feel like they have lots of power.
Think of a bar attached to the flywheel of your engine that is 1 foot long. At full throttle, your engine will produce a force of 18lbs at the end of that bar. At a given RPM, the end of the bar will travel a given distance per second. We know that Work = Force * Distance. The force in that equation is your torque value, and the distance is proportional to your engine’s current RPM. The work value will be an upper-bound on your engine’s horsepower at that RPM. This is why the horsepower chart on dyno graphs usually climbs fairly steadily as RPMs climb, because torque is usually pretty flat and torque essentially describes the slope of the horsepower graph.
In layman’s terms:
-A high-torque engine like an electric motor feels spirited because at low speeds, it feels like it is generating more power than it ‘should’ at that RPM. Diesels also have very high torque.
-A low-torque engine (F1 engine, most motorcycle engines) requires your to rev it quite high in order to get the same amount of power as a high-torque engine. If it has a high enough redline (19000rpm-ish for an F1 engine), it can generate just as much power as a more torque-ey engine and motivate your car just as fast, but around town it might feel a bit strained.
Another interesting tidbit:
The reason that most engines have their torque and horsepower approximately equal is a coincidence because most engines have their redlines around 6000rpm, which is when they just happen to be equal when you’re talking about hp and ft-lbs. You can actually get a decent idea of how high a gas or diesel-powered car’s redline is by its horsepower and torque numbers: if the hp value is much higher than the torque value, it is almost certainly a very high-revving engine (F1: 200ft-lbs, 800hp, 19000rpm). If the torque is much higher, then it is likely a low-revving diesel.
As for the actual thread’s question, maybe you could get an approximation by doing off-throttle glides on a flat road at varying speeds. If aerodynamic resistance grows as a square of velocity but rolling resistance grows linearly, then you could probably solve for a rough approximation.
What would I do off throttle glides with? I have nothing built…
Also, I have seen that, but as I stated earlier I am not sure what to call what I am looking for. But when I say rolling resistance I am not referring to what the wikipedia article does, I am not referring to wheel deformation.
Rolling resistance is essentially determined by deformation, of wheel and ground. Overcoming this resistance has little to do with calculating the torque needed to accelerate your 200lb motorcycle to 30mph as you described. The torque needed depends entirely on the ratio of your driving gear to the diameter of your wheel. It also depends on how quickly you want to do this. You will have to overcome a certain amount of friction and resistance, but up to 30mph these forces will be rather small. Tell me your desired acceleration and gear ratio, and I can tell you your necessary torque.
Try an experiment, put very round, solid metal wheels on something, and roll it along a smooth, level metal surface. Does it take much force to overcome rolling resistance? compare to pushing a heavy car with flat tires over gravel.
You might be confused by the fact that “rolling resistance” doesn’t include drivetrain friction, wheel bearing friction, motor friction, etc. A car with its parking brakes on has a lot more resistance to motion than one in neutral with the brakes off, but that’s not “rolling resistance”.
Trains and other tracked vehicles are more efficient than buses because the metal wheel on metal track interaction produces much less rolling resistance (the wheel doesn’t deform nearly as much) than does a rubber wheel on asphalt road interaction. (Tracked vehicles are also efficient because they don’t need to “generate” power locally but that’s not the point)