I know the drill motor spec sheet tells a Horsepower rating… and I know some teams have done their own testing to verify spec sheet promises…
Has anyone found maximum horsepower ratings for all the kit motors this year? If not all… at least some? i.e. FP, Globe, and Van Door
Gadget,
Again you have to refer to the tables for output power as related to RPM on the motors in question. There is a conversion factor, Watts x 0.001341 = Horsepower. This will allow you to convert from watts to horsepower. You can find more conversions at http://www.watb.net/info/c_factor/power.shtml
Problem is, only the Drill and Chiaphua motors say the Watts Out at max power or the efficiancy of the Watts In
You can calculate the Watts Out at max power yourself, you just need to know how.
Check out my lecture notes: "WRRF motor selection workshop lecture notes… "
It’s under the technical sections in white paper. Here is a link to it: http://www.chiefdelphi.com/forums/papers.php?s=&categoryid=2&action=display&perpage=10&sort=date&direction=DESC&pagenumber=2
I’ve always used 746 W = 1 H.P.
A 400 kW Mercedes engine has 300 H.P. in the U.S.
And since electric motors are really efficient, use Watts Out = > 0.9 times Watts In = Volts x Amps. (The approximation is useful because your input data are approximations too.)
If you have to figure Watts out, you can also use torque times revs :
W - F x d = F x [v x t] = (T / arm) x (rps x arm x 2 x pi)
J = N x m = (N.m/m) x (cirumferences-per-second x length of circumference) = (N) x (length) = F x d = W
To anyone following this thread, or who just may want to know, the following is the maximum horsepower ratings for the 2003 motors.
Drill: .504
CIM: .4562
F-P: .1987
-w/ GB: .121
Globe: .053
Van: .0889
Window: .0074
Information found with the help of Ken and Al’s posts, thanks a lot guys 
And since electric motors are really efficient, use Watts Out = > 0.9 times Watts In = Volts x Amps. (The approximation is useful because your input data are approximations too.)
Errr from what I have read the reverse is true. How efficient is really efficient by your standards??
*Originally posted by Adam Y. *
**Errr from what I have read the reverse is true. How efficient is really efficient by your standards?? **
He used 90% in his formula.
He used 90% in his formula.
’
Yeah that is not really smart to do unless you know your dealing with rare earth magnet motors which generally have high efficiencies. This really is not smart to do with the motors First gives us.
*Originally posted by Lloyd Burns *
**I’ve always used 746 W = 1 H.P.A 400 kW Mercedes engine has 300 H.P. in the U.S.
**
Wouldn’t it be that a 300 kW Mercedes engine has 400 H.P. in the U.S.?
*Originally posted by Gadget470 *
**To anyone following this thread, or who just may want to know, the following is the maximum horsepower ratings for the 2003 motors.Drill: .504
CIM: .4562
F-P: .1987
-w/ GB: .121
Globe: .053
Van: .0889
Window: .0074Information found with the help of Ken and Al’s posts, thanks a lot guys
**
Given the above information, can anyone explain to me how WildStang was able to use 2 Drill and 2 F-P motors, one on each rotating wheel, to go in any direction, without the Drill motors overpowering the F-P motors and throwing the whole bot into a spin.
I guess to go sideways, you could throttle back the Drill motors to match the F-P motors, but then wouldn’t you have .1987 * 4 = 0.7948 effective horsepower versus 1.08 horsepower for 2 Drill motors alone?
Am I correct in assuming that WildStang was able to move faster in a forward direction that in a sideways direction?
The values I posted are the maximum values of horsepower.
Horsepower is a relationship between RPM and Torque. When running the drill with no load and no gearbox, you won’t get .5 HP out of the motor.
Take the following equation: Torque * (RPM/5252) = Horses
Now yes, you may be getting more horsepower to wheel A and B and less on C and D, but that’s OK.
The wheels were “speed matched” where the final stage of gear reduction left all wheels with the same speed by having different ratios.
Wildstang, and others using the same setup, achieved a certain speed and horsepower by matching the speeds while having different torque values to the sets of wheels.
If I’m not mistaken, correct me if I’m wrong please, WS probably had drill putting out ~.4 HP and the F-P’s putting out ~.05
.4 + .4 + .05 + .05 = .9 HP
Maximum Power (in Watts) output is typically calculated as:
Power = (Max Freespeed / 2) * (Stall Torque / 2) * Conversion Factor
Maximum Power (in HP) can then be calculated as:
HP = Power / 746
Really, horsepower is a misleading and useless number. Power output (in Watts) is by far a better and more accurate comparison. When coupled with efficiency, it’s the best comparison.
Horsepower and Watts are the same thing, but HP is an Imperial Unit and the Watt is the SI (metric) unit. SI units are by far better units to use, because you will use less conversion factors (less screw-ups) when moving between Power output and Electrical input. However, remember to mention you are using SI units in all your calculations or you might end up like NASA!
*Originally posted by Gadget470 *
**The wheels were “speed matched” where the final stage of gear reduction left all wheels with the same speed by having different ratios.Wildstang, and others using the same setup, achieved a certain speed and horsepower by matching the speeds while having different torque values to the sets of wheels.
**
That makes sense. I guess if they got in a pushing match going sideways, they could have some interesting differences in the wheel behavior due to the “different torque values”.
*Originally posted by Jnadke *
**Maximum Power (in Watts) output is typically calculated as:Power = (Max Freespeed / 2) * (Stall Torque / 2) * Conversion Factor
Maximum Power (in HP) can then be calculated as:
HP = Power / 746
Really, horsepower is a misleading and useless number. Power output (in Watts) is by far a better and more accurate comparison. When coupled with efficiency, it’s the best comparison.
Horsepower and Watts are the same thing, but HP is an Imperial Unit and the Watt is the SI (metric) unit. SI units are by far better units to use, because you will use less conversion factors (less screw-ups) when moving between Power output and Electrical input. However, remember to mention you are using SI units in all your calculations or you might end up like NASA! **
Thank you. That link is a very cool source of info on motors. Who knows. If I can get some time to study all the tutorials on that web site, one of these days I might understand what our mechanical engineers are doing.
To answer many of the above questions…
The different motors we used were matched to different transmissions so that the wheels were close to match. Any fine tuning could then be done in software. Knowing which side the drills were on then set what side of the robot was used for power, pushing, etc.
Using watts for motor output power helps solve a number of problems in design. You know what the output power is and you can measure the input electrical power, so with a little calculation you can determine efficiency. Power out/power in=efficiency. So what does that tell you? The power lost in the conversion has to go someplace. The result is heat. So in the case of the Bosch motor running at 376 watts out with 672 watts input, efficiency at 56%, then 296 watts go into heat. Yes that’s like three 100 watt light bulbs in each motor doing nothing but generating heat. That is the heat that changes the motor internal resistance, melts the bearing lubricants, detaches the fan and unsolders the motor wires. In many cases in electrical systems, it is very important to get rid of the heat generated in the system. Things like CPU heatsinks and power supply fans in your computers prevent that heat from building up to point where the physical properties start to change. In solid state electronics, the heat generated internally can cause more current to flow, which in turn causes more heat, and so on. The tech term is “thermal runaway” and many devices are designed to shut down when the temp gets too high. The three terminal regulator, 7805, will cutback on current or shut down completely if it gets hot, for instance.
*Originally posted by Al Skierkiewicz *
** … In solid state electronics, the heat generated internally can cause more current to flow, which in turn causes more heat, and so on. The tech term is “thermal runaway”… **
Good info! The same thing is happening in our motors: the magnetic field generated by typical permanent and electromagnets is inversely proportional to temperature. So as a motor heats up, it needs more current to produce the same amount of power, and at 56% efficiency, that means more heat, which means less power… You get the idea.
Compounding this is the fact that the motor relies on a fan attached to the armature for internal cooling, which makes its effectiveness proportional to speed.
These factors, along with battery drain, explain why a robot which might have good power in testing and early in the match, might have a hard time getting up the ramp after a lengthy pushing battle.
The strength of the magnetic field of a motor going down is a direct result of the internal resistance of a motor skyrocketing.
As we all know, the force of a magnetic field is proportional to current as:
Force = IBL = I^2 * C
I = Current
C = N * L * u0 * Xm / l
I’ve shortened this equation because the things in the constant C, are never changed during the operation of the motor. Some are, but they are very small. If you want to know what this equation means, or how I got it, or think it is wrong, PM me first. If you still think it’s wrong, then you may argue your point here. Now, the current draw of a motor is related by the equation:
Current = Voltage / ( R + Rz )
R = Internal Motor Resistance
Rz = Motor Impedance (Resistance to Current)
Now, as a motor approaches stall, it’s impedance becomes 0, because the windings aren’t moving and there is no magnetic field to resist a change in current, so the current draw of a motor can be accurately represented as I = V/R.
R of Drill Motor = 94.5 milliOhms
R of Chiaphua = 105.3 milliOhms
Now, V, for the most part, remains constant at 12V. The only thing that is able to change is the resistance of the material. So, lastly, the Resistance of a Material, R, is affected by temperature according to the equation:
R = R0 + aT
R0 = Resistance of alloy at 0 Degrees C
a = Material-Dependant Constant
T = Temperature
As with every other equation above, there is only one varible that is really changed, and that is T, Temperature. All the other symbols are constants.
As you can see, comparing this equation with the first equation, any increase in temperature is SQUARED as a magnetic field decreases. Furthermore, as said above, it is inversely proportional, in that:
Force = C / (Temperature)^2
Now you might just think twice about how important it is to cool your motors…