Big Gear, Low Thottle vs Little Gear, High Thottle

[summerof95]

I'm curious which of these two situations would make most efficient use of available power.

Larger pinion gear, 3/4 throttle.
Smaller pinion gear, full throttle.

Assuming that every other aspect of the two situations is a constant, including the vehicle's velocity, which vehicle will deplete it's battery first?

[josh6575]

Depend on the amps drawn. If the amps are the same, then they will run for the same amount of time. The higher geared one will go further.

[Exocaged]

If you never need that last 25% of wheel speed the larger pinion gives, then do the smaller one and go have fun. Also check temps as your motor/esc may choose for you.

[summerof95]

Josh, if they run for the same amount of time, with the same velocity, how could one go farther than the other, since both time and displacement are factors of velocity?

Lets use speed instead. If they both go 10 mph for one hour, in a straight line, they both would have traveled ten miles, Yes?


So lets go back to amps. Would they draw the same amps?
Draw is related to load, and the second example would have lower load. Would it be more efficient?

[dna4engr]

It is directly related to current draw on the batteries. My gut says the lower pinion gear will yield longer run times because I believe it will yield lower current draw on the battery. In practice I've witnessed this occuring, however the lower gearing always meant slower speed. Since your experiment is utilizing 75% throttle for the higher gearing it will depend on the efficiency of your esc at that point vs. 100%.

If you could test with amp meter, the two scenarios, you will have your answer. If you don't want to test, just go with the lower gearing.

D

[Olle P]

It's all about efficiency.
Each motor has a peak in efficiency a specific load, where "load" is a percentage of stall torque at the voltage currently fed.
If the desired run speed is given it's a matter of finding the gearing that lets the motor operate at peak efficiency at that wheel speed.

For cheap brushed crawler motors maximum efficiency is at ~20% load, whereas for brushless motors it's at ~5% load.

[summerof95]

So typically peak efficiency is 20% load, meaning 20% of the motors stall load, right?

Is there a clever way to measure motor stall load?
Fish scale and a ruler?

[Olle P]

Originally Posted by summerof95

So typically peak efficiency is 20% load, meaning 20% of the motors stall load, right?

Correct.

Originally Posted by summerof95

Is there a clever way to measure motor stall load?
Fish scale and a ruler?

Should work, at least in theory, although I suspect the scale is way too course to measure these low forces. A scale for letters seems more appropriate.
* The torque does differ a bit depending on the rotor's orientation relative to the can.
* Do use low voltage! The current will only be limited by the motor winding. The torque is proportional to the current, so stall torque is also proportional to the voltage.

But as the others said: Current drawn from the battery (as opposed to "fed to the motor") is a good measurement of actual efficiency.
Using a logger like for example Eagle Tree eLogger you can try different gearings and find out which provides the best overall efficiency.

[JohnRobHolmes]

At the same vehicle velocity the small pinion setup will have higher efficiency as the amp load is lower and voltage higher for the same wattage. A motor always is more efficient when run faster with more geardown, until motor speed is so fast that core losses dominate.

It's the reason I've told people to volt up and gear down for a decade straight!

[MountainStorm]

If you set up a brushed high RPM motor with a smaller pinion geared to run the same MPH at peak efficiency as a brushed lower RPM motor using a larger pinion running at peak efficiency which set up would deplete a 3S battery first? I assume the lower wind motor is more efficient. This over looks what the vehicle needs to do besides run down a battery pack but I wonder if I am correct in that assumption?

[JohnRobHolmes]

We can assume the higher rpm setup will have an efficiency advantage as long as the RPMs are not so high that core losses have overtaken copper losses. You are basically comparing the same thing as the original question but using different wind of motors instead of different phase voltage, two systems with the same wattage requirements where one motor spins faster than the other.

Any time we can increase RPM to reduce amps for a given load it will increase the motor efficiency until switching losses in the core go wild. This RPM limit can be found by running a motor no load and plotting the current against RPM. At a certain speed the current needed for further rpm increase does not follow a linear line. From experience I know this is right at 50,000 rpm for a 3 slot or 5 slot using standard laminations with two magnetic poles. You can also derive this point from the switching rate and the lamination grade, 400hz is a typical recommended max point for the laminations and 400hz x 60 seconds x 2 poles = 48,000 rpm

[summerof95]

A motor is intended to convert electrical energy into rotational force. A portion of the energy is always "lost," and is generally converted into heat.

If more efficient setup means more energy is successfully converted into rotational force, that must leave less energy left to convert into heat.

Meaning that even though the motor is spinning faster, it will actually run cooler, right?

[JohnRobHolmes]

Power as watts= amps x volts
Power as watts= N*m torque x RPM
Torque is generated with amps
RPM is generated with volts

If we can increase motor rpm to generate the same power, amps will be lower. Amps make heat and voltage drop inside the wire * amps = watts of waste heat in copper loss. Reduce the amps and output efficiency increases, while waste heat generated in the copper is reduced.

[Olle P]

Physics and math to follow below...

Originally Posted by JohnRobHolmes

... A motor always is more efficient when run faster with more geardown, until motor speed is so fast that core losses dominate.

Originally Posted by JohnRobHolmes

Any time we can increase RPM to reduce amps for a given load it will increase the motor efficiency until switching losses in the core go wild. ... At a certain speed the current needed for further rpm increase does not follow a linear line.

I think this is a bit on the extreme.

Assuming we operate well within the linear region the following apply (somewhat simplified, not accounting for factors like battery voltage drop under load and non linear bearings):
("Load" below is a fraction of stall torque, given as a value in the [0 1] range.)
Power drawn by the motor:
[Feed voltage]*(([Stall current]-[No load current])*[Load]+[No load current])
(Voltage * Current)
Power provided by the motor:
[Feed voltage]*[Kv]*(1-[Load])*[Stall torque]*[Load]
(Speed*Torque)

Interpretation:
* The power drawn is thus a simple straight line function of load (Pd = k1*[Load]+P0), starting above zero.
* Power provided boils down to Pp = [Load]*(1-[Load])*k2 = k2*([Load]-[Load]^2).
It peaks at 50% load and is zero at the ends.
* Power efficiency is provided/drawn and thus
([Load]*(1-[Load])*k2)/(k1*[Load]+P0)
Maximum efficiency is located somewhere between 0% and 50% load, exactly where is dictated by the relationship between the stall current and the no load current.

Example:
Relevant data relatively often found for motors:
* No load current (I0) at a given voltage (U0).
* Motor (winding) resistance R.
Then peak efficiency, as given by some basic calculus applied to the formulae above, is reached when the load equals
2*I0*(SQR(1+(U0/R-I0)/I0)-1)/(U0/R-I0)
Notice that we don't get to know what power (or torque) the motor will deliver!

While driving straight ahead at a constant speed on level and flat ground the load should be below the maximum efficiency level in order to leave room for the extra current (power) required for acceleration and driving in other terrain.
Some very high power motors can't even handle the current required at peak efficiency on full throttle, at least not for long!

I think my point is that "gearing down" isn't always better, at least not for efficiency.

[Dialed]

Originally Posted by summerof95

I'm curious which of these two situations would make most efficient use of available power.

Larger pinion gear, 3/4 throttle.
Smaller pinion gear, full throttle.

Assuming that every other aspect of the two situations is a constant, including the vehicle's velocity, which vehicle will deplete it's battery first?

Is this brushed or brushless motor you are speaking of?

[josh6575]

Originally Posted by summerof95

Josh, if they run for the same amount of time, with the same velocity, how could one go farther than the other, since both time and displacement are factors of velocity?

Lets use speed instead. If they both go 10 mph for one hour, in a straight line, they both would have traveled ten miles, Yes?


So lets go back to amps. Would they draw the same amps?
Draw is related to load, and the second example would have lower load. Would it be more efficient?

Try it and get back with us. Of course velocity X time is going to equal a certain distance.

[summerof95]

Olle and Holmes, thanks for the info!

I noticed Kv popping up in those equations, which is determined by unloaded RPM/Feed voltage, right?
So if the RC4WD 27t motor is 15000rpm @ 7.2v, that would give it an approximate Kv of 2083.3, right?
Now, I don't typically see Kv used in the context of brushed motors.
Is there a reason for this? Tradition?

Originally Posted by josh6575

Try it and get back with us. Of course velocity X time is going to equal a certain distance.

Yes, trying it for myself would have been the most expedient way to test it. But until Olle P has pointed out the Elogger, I never knew such a thing existed, and hadn't known that this could be so easily measured.
Regardless, there are certain factors that could potentially skew any real-world testing of this. For instance, the average amp draw measured would likely include the period of acceleration, which would likely favor the higher ratio gears. I suppose this could be minimized by driving the longest possible distance, but I simply lack the space to drive in a straight line for such a distance. Turning isn't really an option; with all three "locked" differentials, turning would introduce a significant increase in drag, again, likely favoring the taller gears.

But with the number of dedicated folks on this forum, I figured someone would already know the answer to this, avoiding the problems of a physical test.


As for velocity, displacement, and time, one can always be determined by the other two. That was my point: your example of a constant velocity and consent time producing two different displacements is impossible, no matter what motor/gearing combination is used.

[Olle P]

Originally Posted by summerof95

... I don't typically see Kv used in the context of brushed motors.
Is there a reason for this? Tradition?

I have the impression that Kv for brushed motors isn't a constant but rather a function of feed voltage.
Should be good enough for a rough estimate of load for maximum efficiency though.

Originally Posted by summerof95

For instance, the average amp draw measured would likely include the period of acceleration, ...

Using the eLogger you get data for the entire period from first pull on the throttle to power off. It's visually represented by graphs, so you can correlate the graph to your drive.
Then you can choose any time span you see fit within that period and get all the data you want (max, min, average) for that sub part only.

It looks like this (sampled from my crawler years ago):

You have the battery voltage in purple and current drawn from the battery in orange.
* The first bracket indicates me holding the crawler so the wheels spin freely.
* Second bracket is driving straight back and forth on the floor.
* Third bracket is driving around in circles on the floor. (You can see the current fluctuate as the front tires get and lose grip.)
* Last bracket indicates fast accelerations back and forth.

[EeePee]

JRH got some sample KV readings from brush motors years ago.

Motor speeds (taken from random and averaged sample @ zero timing & 3100rpm):
10t = 6526 rpm/volt
15t = 4317 rpm/volt
17t = 3753 rpm/volt
20t = 3216 rpm/volt
21t = 3082 rpm/volt
23t = 2790 rpm/volt
25t = 2588 rpm/volt
35t = 1565 rpm/volt
45t = 1210 rpm/volt
55t = 987 rpm/volt
65t = 837 rpm/volt