Phisic's Problem Here !!

[TiagoSilva]

Hi, i found myself a phisic's problem, i'm choosing an motor to a robot / rc-car, i only have one equation to get the torq (Newtons) metric weight (Grams/Kgrams), or i think i have ...

I keep finding mNm, mN.m and Nm, mostlly at farnell and rs ...
Wich one is it ?
Wich is bigger, i guess that Nm is bigger than mNm, but i keep thinking that something is not right here, i just don't know what ...

Usually it's N (1:1 ?) and Nm (1:1000 ?), i guess ... then why the mN ???

Where can i get some more info about it, or a free ebook, something that helps me to know what torq i need to pull some weight.
Ex. 1,5Kg keeping at least 3500RPM (Too mutch ?) ...

I get 32KM/Hour from a 50Mm weel at 3500RPM.
32,986... KMh = ( ( 3500RPM * ( ( 50Mm * Pi ) / 60 ) ) / 1000 ) * 3600

This i know : N = Kg x Gravity, then Kg = N / Gravity, right ? Well, i know that much ... LOL !!! I guess i'm lame ...

Then if i have a torq of 0,15 Nm,
1,800Kg = 0,20 x 0,9.

Is it prudent to use Step-Motor ? How can i know the max RPM of a Step-Motor, since the vendors don't show that info on them ...

Is this right ???

Help most apreciated
Best Regard's Tiago.

 

[Papabravo]

Stop!
Use the same units consistently until you become familiar with the relationships.

F = ma  --> Force is mass times acceleration

T = I * alpha  --> Torque is Moment of Inertia times angular acceleration

Don't mix these two things up. That way lies disaster.

In the MKS system

Units of force --> Newtons
Units of mass --> Kilograms
Units of acceleration --> Meters/second/second

Units of Torque --> Newton-Meters
Units of angular acceleration --> radians/second/second

Radians have no dimensions, so can you tell me what are the MKS units of Moment of Inertia? The units have to be the same on both sides, or else it is not a valid equation. Dimensional analysis will reduce all units in to fundamental ones which are units of length, of mass, and of time. So

1 N = 1 Kilogram-Meter/second/second = 1 Kilogram-Meter/sec^2

 

[TiagoSilva]

Neat , well, i didn't knew this mutch lol, but what i need to know is how mutch Units of Torque i need to carry a given Weight (in Kg), that way if i knew the Newton-Meters needed to carry for example an RC Car that weight's 2Kg i could choose the needed Motors.

Since i'm not that expert in physic's i can be making the wrong questions ...
I hope that this reply shed some light on you of what i whan't make with theese calculations.

 

[dknguyen]

That's a tricky topic since it depends on terrain, friction between ground and wheels, wheel deformation and other things. You can't measure the friction of wheels very easily with every single possible type of terrain. So you have to do a rough upper limit estimation.

1.
F = force of rolling friction +force of gravity along surface
F =uN+S

F= force required
u = coefficient of rolling friction
N = Normal force
S = force that causes rolling down the hill caused by gravity
________________________________
If you are trying to wheel up an slope with an inclination of D degrees (or radians) and a mass of m, then the formula works out to be:

F = u(Mg)cosD + (Mg)sinD

F= force required
u = coefficient of rolling friction
M = mass of vehicle
g = acceleration of gravity
D = inclination of slope

You can also replace (Mg) with W if you already have the weight (a force)already as might be the case if you choose to use imperial units (bleck!) instead of metric.
==================================
2.
After you find F, the total required total force exerted by the wheels, find the force required by each wheel (assuming the weight of the robot is evenly balanced over each wheel).

Fwheel = F/(# of wheels)
==================================
3.
THen you can find the torque needed by each motor in each wheel:
T = Fwheel*(radius of wheel)
=================================
S is the force that causes the vehicle to roll down the hill. It is the component of the force of gravity acting parallel to the surface of the ground. On flat ground, this force is zero (gravity does not cause the vehicle to roll). On a slope this force is greater than zero, but less than the weight of the vehicle. On a cliff, this force is the weight of the vehicle (ie. the vehicle is falling straight down).

The normal force, N, is the force the ground must perpindicular to it's surface to stop the vehicle from falling through. On a flat surface this force is equal to the weight of the vehicle, and is less on a slope. On a vertical cliff (90 degrees) this force is 0. The normal force and u, the coefficient of friction dictate how much force is required to get the vehicle rolling.

To roll up a hill, you must be able to over come the force of friction (force of friction = coefficient of friction x normal force) and you must also overcome the part of gravity causing the vehicle to roll down the hill.

For the coefficient of friction (rolling friction more specifically) just pick a good number between 0 and 1. 0 means no friction. You can't measure the friction between your wheels and every possible surface you plan to ride on. 0.03 is a good one for flat even ground. Maybe 0.1 on rougher ground. I wouldn't use much higher than 0.3 unless your wheels are globs of mushy rubber.

u, the coefficient of rolling friction represents the percentage of the vehicle's weight that you need to apply to the vehicle to get it rolling on FLAT ground. For example:
u =0.03 -> force of 3% of vehicle weight needed to start rolling on flat ground
u = 0.1 -> force of 10%
u = 0.3 -> force of 30%

Setting D, the angle of inclination to zero degrees will assume you are on flat ground instead of a slope. 90 degrees will assume you are over the edge of a cliff. Anything between 0 and 90 degrees is a slope.

Understand?

 

[TiagoSilva]

Yes, the (total weight) = ( car weight ) + ( car gravity weight ), and 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 …
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.

Let’s try to solve this with this data :
Car Weight = 2Kg ( 2000 g )
Number of Wheels = 4
Slope Max Inclination = 0,30º



F = u(Mg)cosD + (Mg)sinD
F = 0,25 ( 2000 g ) cos 0,30 º + ( 2000 g ) sin 0,30 º
F = 500 + 10,47 (…)
F = 510,47

Fwheel = F/(# of wheels)
Fwheel = 510,47 / 4
Fwheel = 127,6175

T = Fwheel*(radius of wheel)
T = 127,6175 * 0,0075 Metres
T = 0,95713125 N


Guessing this is correct, I think i need a motor capable of 1N per Metre right ?

Thanks a lot , you have been quite helpful.
Again, thanks a LOT !!!

 

[dknguyen]

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).
================================
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).

 

[TiagoSilva]

Well, as i said i'm somewhat new to this, and thus, the flaws like using the wrong names ... yet you knew what i was saying, thank God, i also have this equation to get the speed the object goes, and here it is :
KM/H = ( RPM * ( ( ( Wheel Diameter in mm * Pi ) / 60 ) / 1000000 ) ) * 3600
This was made from this one :
mm/sec. = RPM * ( ( Wheel Diameter in mm * Pi ) / 60 )

Then i know that if i need more speed i need a bigger wheel or a faster motor, but bigger wheel's mean more torque needed ...

So, what if i whant to make the car run for let's say 30KM/H, i need 2150RPM with a wheel of 75 mm ( same wheel size used above ).

Will that changes the torque needed ???
( i was hopping not since the torque is the Force and RPM the times the wheel turn )

 

[dknguyen]

The equation looks right..but, BUT, and this is a big but in your first post with calculations, you used a wheel RADIUS, not diameter of 0.0075m=7.5mm.

Did you make sure to compensate th for the 7.5mm? Because when I did the equation for the torque per wheel, I was using 0.0075m as the radiusrather than the radius. If 0.0075m is actually the diameter, you can reduce the required TORQUE calculated by half (the required force at the edge of wheel stays the same, but since the wheel is smaller by half, you need half as much torque).

WIth your speed requirements, what you now need to do is to find a motor that can put out the required amount of torque WHILE spnning fast enough. This means you need a motor of sufficient power (you can adjust the motor gearing or wheel size to convert speed-to-torque or torque-to-speed). So now you should be looking for motors that put out that much force, or more force, while still being able to spin fast enough (remember, motors spin slower if they have to apply more torque).

Torque is only half the story. Speed is the other half. Power is what brings the two together. If you have a motor of enough power, but of insufficient speed or torque (a motor with enough power, but insufficient speed/torque means that it has an ambudance of torque/speed.). Then you gear it down or make larger or smaller wheels or whatever to shift the balance between torque and speed in order to get what you want.

Your last sentence, you have a word mixed up, POWER is the
torque x angular velocity (RPS in radians)
or you could also say power is
torque x angular velocity (RPM/60 in radians)

So a simple answer your question, is no, torque does not change to travel at the speeds you want, BUT motor power must increase.

PS. Why on earth would you need something with 7.5mm wheels to travel at 30km/h?!! That's 8.3m/s and is VERY fast for a large robot already, let alone a small one. How will it navigate? 8m/s? That speed is like...the speed a fly might fly at. Have you measure 8m on the ground and tried to cover that distance in 1s? That's right around sprinting speed. Do you have square floor tiles anywhere? Most of those are 1 foot by 1 foot. 8.3m is 27 of those squares! With such small wheels and light weight, it will either smash into big bumps and destroy itself or ROCKET over small bumps and crash.

 

[TiagoSilva]

Oooops, and a big one that is ...
I mislead you, i'm very sorry, but i was wrong all the time too, it is not 0,0075m but 0,075m as in 7,5cm ... Damm ... , well, yes i was using it as diameter and not as the radius ... Sorry again ...

Then, here is my update,
F = ( 0,25 x 2 ) x cos 45º + ( 2 x sin 45º )
F = 1,76

Fwheel = 0,44 ( 1,76/4 )

T = 0,44 x 0,0375 ( now as radius, diameter 7,5cm ) = 0,165N = 165mN

One more thing, does dividing F by 4 wheels implies the need of a motor per each wheel ? since the force if divided and you mentioned that the weight needed to be equaly distributed by the 4 wheels ...
Will i need a Motor of 165mM per Meter ???

Thanks again and sorry for the mess ... i was also thinking wrong ...

 

[dknguyen]

Yes, dividing by four implies 1 motor per wheel with 4 wheels. If you only have two motors then you divide by two. The motors could be driving one wheel each, or 5 wheels each, or whatever. I should have said

Fwheel = F/(# of motors)

You can also change the # of motors used in the calculation in some cases. Like if I had a 6 all-terrain wheeled robot, I might say 3 is the # of wheels if I want to account for the fact that not all 6 wheels will not always be on the ground at the same time, but at least 3 wheels always will be. This way I end up with a very conservative number. So if all 6 wheels were on the ground, I would have even more torque than I need.

You can see how if you have a 3 wheeled vehicle, but one of wheels has zero weigh over it, there will be no normal force between the wheel and the grond and therefore no friction, so the wheel won't do anything. THat is the kind of thing I mean when assuming that the weight on all the wheels is equal- each wheel is working equally.

In your calculation you are forgetting the variable g for gravity that each term should be multiplied to.

g =9.81[m/s^2]

The units for torque T is not based on Newtons (that is force), it should be based on N-m or Newton-meters.

I get Fwheel = 4.33N per wheel assuming 4 motors, and T = 0.163N-m = 163mN-m = 16300mN-cm.

 

[TiagoSilva]

Nice Thanks, then for 4 wheels i need at least 2 motors and at most 4 is that it ? Thanks again.

Best Regards,
Tiago Silva.

 

[dknguyen]

Usually two motors is the minimum (one for each side). Otherwise you can have any number of motors you want- one for each wheel, one for every two wheels. You could have a 6-wheeled vehicle with only 2 motors (one per side) or with 6 motors (one per wheel). It doesn't matter.

Another option to one motor per side is one drive motor to drive the both backwheels identically at the same time and a steering motor out front. But, of course, the steering motor is a bit fancier.

The more motors you have, the weaker each motor can be (remember to take into account that not all wheels might always be on the ground especially if you are climbing stuff, so some motors might be working harder than calculated if you did not account for this.).

So, yeah, two motors is the minimum. The maximum is whatever you want...you could have 5 motors per wheel if you really wanted by using a belt or chain drive. But yeah, practical max is 4. Good luck.