Choosing the right motor

[Kars]

Hi there

I want to build a robot, and for that robot i need motors. I don't want servos, they aren't fast enough (they are about 60rpm)
I need +- 3kg-cm torque, and about 300 RPM. I've been searching for some motors and i found this one
They look ideal to me, but when i look at the specifications in the pdf file tehy say that the 'Kt' is 130Mn-M/A and the 'Kv' is 71rpm/v
What is 'Kt' and "Kv'?

Did i choose the right motor?
Do u guys have some alternatives for motors?

Thanks in advance,
Kars

[Admin]

check this tutorial out:
http://www.societyofrobots.com/mechanics_dynamics.shtml

It will tell you if your motor can do what you want.

So robot speed isnt only a function of motor rpm, but also a function of wheel diameter. Servos are really strong and slow. Use a wheel twice as big, and your robot will move twice as fast.

'Kt' is 130Mn-M/A and the 'Kv' is 71rpm/v
What is 'Kt' and "Kv'?

Kv says that for every volt you give to the motor, it will go about ~71rpm faster. So if you give it 6V, it will go about 426rpm - waaaay to fast for most robots! You probably dont want more than 100-300rpm, depending on many different factors.

Im not sure what Kt is, but Im guessing Mn is millinewtons, and A is current. M might be meters. (check the datasheet, it should say!)

So if my guess is right, then it is .13 Nm/A, meaning your motor will apply .13Nm of torque for every amp of current.

[Ro-Bot-X]

How do you transform N*m in kg*cm?

[Admin]

How do you transform N*m in kg*cm?

ehhhhh kg is mass, while N is a force . . .

if you multiply mass by gravity, you can get a force.

[Ro-Bot-X]

Ok, Torque is a force right? Then is corect to say the motor has a max torque of say 15mN-m. But how is that converted to kg-cm or oz-in? Just by multiplying it by gravity (9.81m/s^2)? And of course multiplying by 100 to account for meters to centimeters transformation...

[Admin]

Torque is a force right?

wrong - torque is force * distance

torque can also be mass * gravity * distance

[Ro-Bot-X]

If torque=mass*gravity*distance=force*distance,
we can say in this case that force is mass*gravity?

In other words, to transform from 15mN*m in kg*cm:
(15mN=0.015N and 1m=100cm)

0.015N*9.81m/s^2*1m=0.14715kg*1m=14.715kg*cm

Is this correct or I still get it wrong?

[Ro-Bot-X]

I just got this and I am posting it here, for others to see:

Torque Conversions:
1.0 kgf-cm = 0.098 Nm = 98 mNm
= 13.887 oz-in = 0.867 lb-in

I am trying to find a couple of motors for my big robot and I am using admin's RMF calculator. That's why I ask these questions, because I don't understand exactly how to use it with the metric system. I am looking at a motor and I feel it is not powerfull enough, but in the calculator it is like 20 times more powerfull than I need...

I have all this data filed in:
mass=10kg
accel=0.5m/s^2
incline=5deg
velocity=1m/s
efficiency=0.8
wheel rps=1.5
gravity=9.81m/s^2
It calculates RMF = 2.695683

The motor has 200 RPM and max torque of 18.38 kg*cm
If I put in torque = 18.38 and rps = 3.33 I get RMF = 61.2054...
If I put in torque = 0.1838 and rps = 3.33 I get RMF = 0.612054

Which is correct?

[Admin]

A kgf is kilograms as a force, meaning its multiplied by gravity.

So 1 kgf is 9.81 newtons (gravity is 9.81 m/s^2), or 9810 mN. If distance is 1 meter, then torque is:
9810mN*m

Now to convert that to lb*in, I usually use google to do the conversion for me

[annoyin_kid]

The motor has 200 RPM and max torque of 18.38 kg*cm
If I put in torque = 18.38 and rps = 3.33 I get RMF = 61.2054...
If I put in torque = 0.1838 and rps = 3.33 I get RMF = 0.612054

Which is correct?

just a quick question, what is RMF???
without knowing what it is i rekon that both calculations should be right because in the second one you divided the torque by 100 and the answer was also divided by 100 aswell but this is only right if the values are directly proportional.

hey admin just a quick question, in 'scientific calculations' do u guys in the US use the imperial system or the standard international units.

[Admin]

The motor has 200 RPM and max torque of 18.38 kg*cm
If I put in torque = 18.38 and rps = 3.33 I get RMF = 61.2054...
If I put in torque = 0.1838 and rps = 3.33 I get RMF = 0.612054

You made a simple mistake. You need to convert 18.38 kg*cm to kg*m, for 1.838 kg*m

So the motor RMF is 6.1267, while your required RMF is 2.7.

Your motors arent actually that great, as your robot (with those specs) still cant go up a slope more than 16 degrees.

[Admin]

just a quick question, what is RMF???

http://www.societyofrobots.com/mechanics_dynamics.shtml

hey admin just a quick question, in 'scientific calculations' do u guys in the US use the imperial system or the standard international units.

Engineers in the US use both. Older engineers tend to only use the imperial system. A lot of engineering software and machining equipment only uses imperial.

Personally, its soooo much easier to think in imperial units (because I used it for the first 20 years of my life). But when it comes to calculations, SI is much easier to work with.

[Ro-Bot-X]

You made a simple mistake. You need to convert 18.38 kg*cm to kg*m, for 1.838 kg*m
So the motor RMF is 6.1267, while your required RMF is 2.7.
Your motors arent actually that great, as your robot (with those specs) still cant go up a slope more than 16 degrees.

1. If 100cm = 1 meter, then why is kg*cm divided only by 10 to get kg*m?
2. The motors are not the best ones, but the next stronger motors would have 355kg*cm torque and no encoders, consuming lots of amps . But there are 2 motors on a robot, won't that make a double RMF?

[Admin]

oops, you were right when you said:

If I put in torque = 0.1838 and rps = 3.33 I get RMF = 0.612054

But there are 2 motors on a robot, won't that make a double RMF?

yeap, the RMF is doubled because the torque is doubled.

2.7 RMF required, but with two motors you are only getting like 1.2 RMF

so you are right, your motors arent powerful enough . . . you could of course reduce expected velocity/acceleration (which by the way I think is a bit too high for a robot - unless its a battlebot or racecar).

[Ro-Bot-X]

Thanks!
What would be a reasonable max speed for a 1.2 meters tall indoor/outdoor robot? I intend to use it at about 50-70% of max speed.
I thought people walk with a 6 km/h speed. That means 1.6 m/s. So probably the max speed should be 0.8 m/s for outdoor and maybe 0.4 m/s for indoor?

[Admin]

So for speed . . . the real question is, how fast can your computer process sensor data and output it as motor commands?

Then the question of terrain, acceptable vibration on your electronics (hard drive especially), tendency to flip over on sharp turns, etc . . .

Assuming your robot is similar to Eric (the next upgrade?), Id say perhaps .3 m/s. Acceleration probably .3m/s^2 too.

But again, it has a lot to do with Eric Jr's ability to sense and avoid obstacles.

[Ro-Bot-X]

Yes, it's going to be Eric the Second. 

Because I will move to Canada in the spring, I will have to sell some of my robotic parts (it's too expensive to get them transported over there, I allready had them transported from USA to Romania once...) and I was looking around for better replacements.

I want to build a ballancing robot, so when is the tutorial coming up? I don't have the math behind me to do it myself (at least not alone) especialy the software (Kalman filters...)!

I understand the sense-compute-command part, speed will be adjusted accordingly, but the max speed has to be calculated beforehand. The robot may never run at that speed, it will get to 90% only in the best conditions, like smooth terain, no objects sensed by the short and long range sensors... Anyway, a ballancing robot has to accelerate to get to a stop. Also, a ballancing robot it is hard to tip over because it is dynamicly ballanced and will bounce off. The conclusion that i got is: a small diameter base and tall robot HAS to be dynamicly ballanced or it will easily tip over.