All right, so you've read the basics on motors (if you haven't, check here). Now you know roughly how they work, and some attributes about them. But chances are, you aren't going to find a motor that perfectly fits your needs normally. There are ways to improve motors' performance, and one way is by messing around with the voltage. However, there are various things to consider when choosing a voltage to run a motor at, and these things affect the entire electrical system of the bot. So, before you start experimenting with voltages, you ought to learn a little about the math in here. But prepare yourself: we'll be dealing with a number of numbers very shortly.

I've said earlier (I think) that the higher the voltage you run a motor at, the faster the shaft spins (and the less voltage you apply, the slower the shaft turns). For some motors, a small change in voltage means a great change in RPM, for others, a small change in voltage means an average, small, or even almost unnoticeable change in speed. But whatever the case, this change is at a constant rate. For a particular motor, the RPM is always a certain multiple (or fraction, as the case may be) of the voltage applied. In algebra, this is called a "direct variation" and is shown as y=kx. But people who deal with motors are as sick of that as you probably are, so they re-wrote the equation to reflect the situation: RPM=(Kv)(V) where V is the voltage you chose to apply, Kv is a constant that remains the same at all times for your particular motor (but varies from motor to motor), and RPM is the RPM you will get at that particular voltage. But let's put this into use!

Suppose you have a motor which normally runs at 9V, and has 81 RPM at that speed. You can either plug in these numbers into the above equation and maneuver them around to solve for Kv, or you can maneuver the equation and then plug in the numbers. I prefer the latter, but that's just me. So, Kv=RPM/V (and, by the way, RPM/volt also happens to be the unit for Kv, making the equation easier to remember). Kv=(81/9), Kv=9. So, this motor has a Kv constant of 9 revs/min/volt. Now, suppose you want 120 RPM. You now know Kv, so plug in your numbers: V=(RPM)/(Kv). RPM is the desired RPM in this case, and V is the voltage necessary to reach that. So, V=120/9, or 40/3 (about 13.33 volts). Obviously, this isn't a common voltage, so probably you'll have to settle for a common voltage that is closest to this estimate. You might get a 11-cell NiCad pack (13.2 Volts) and settle for 9(13.2)=118.8 RPM, or if the RPM is not that important, you may settle for 12V (10 cells NiCad, 8 alkaline), a very easy-to-find voltage, and only get 108 RPM. You will have to decide which matters more to your design. (A thing to note: the RPM from this equation is ideal, it's the "no-load speed", the speed where there is no torque required (this never happens, though, since even spinning the shaft alone requires torque). When the motor must exert force, the top speed lowers. I'll derive how to calculate that later.) Anyway, that's not too hard, is it? You just have to get used to tying algebra into it.

The stall amperage is an important thing to know about a motor too: the max amount of amperage it will give at a certain voltage. Back in the day, when electricity was being discovered, a scientist named Ohm discovered a law called Ohm's Law (strangely enough). This law is: V=IR, I=V/R, and R=V/I. V is voltage, I is current (amps), and R is resistance (ohms. Guess who this was named after?). This law applies to motors, but only in certain cases. Due to some complex (or at least above my head) electrical situations, Ohm's Law is not that simple while the motor is spinning. But when the motor is stalling, as the torque required exceeds how much the motor can provide, the motor turns into an incredibly large resistor. So, in stall situations, Ohm's law DOES apply. This means that the stall current will equal the voltage applied divided by the resistance of the motor. Here's the important part: V and I can change, but R is constant. So, you find R, you can solve for the stall current at any voltage. Setting that up: R=V/I, or resistance=voltage applied/stall amperage (amps) at that voltage.

Suppose you know that at 9V, the earlier motor had 27 amps stall. You want to solve for the resistance. So, R=V/I, R=9/27=1/3! Now, what is it at 12V?? You know R and V, so solve for I. I=V/R. I=12/(1/3) or 12*3. I=36 amps at stall. Woo-hoo! Coolness.

Now, torque also has a constant: Kt. Kt can be tied back to Kv, as Kt times Kv for every motor is equal to 1352. It's a fact of life. However, torque is not directly related to voltage times Kt. Torque is related to amperage being drawn at any time. In fact, torque is not the variable: amperage relies on how much torque is being pulled from the motors at a given time. So, if I were to make an equation for amperage, it would look something like this: A=(T)/(Kt). And from this, knowing the stall torque, Stall Amperage=Stall torque/Kt. Kt is in units of oz-in/amp. THIS IS IMPORTANT! You may get used to using in-lbs as units, since the bigger you go, the more torque there is, but oz-in is the unit to use for your torque in this equation. To convert from in-lbs to oz-in, just multiply by 16 (16 oz=1 lb).

So, going back to our original example: we found the Kv of our motor to be 9. Since we know that Kv*Kt=1352, set the equation up. Kt=1352/Kv. 1352/9=150.2=Kt. So, the motor has a Kt constant of 150.2 oz-in/amp. We found that the stall amperage is 27 amps. So, to solve for stall torque, Stall Torque=Stall Amperage*Kt. S.T.=27*150.2=4056 (not rounding what Kt originally was) oz-in. This is the same as 253.5 in-lbs, or about 21 ft-lbs. Dang! Then, what about at 12V? The stall amperage has increased to 36 amps, so Stall torque=S.A.*Kt. 36*150.2=5408 oz-in, or 338 in-lbs, or about 28 in-lbs.

What about horsepower? Surely you've heard about that: most car engines are rated by it. This is a good segway into Power Transmission. Horsepower is how much power you get out of your motor. A motor with 2 HP will provide more torque (twice as much, I think) at the same RPM as a motor would 1 HP would. This requires gearing down, but like I said, I'll cover that in Power Transmission. Horsepower takes both no load RPM and stall torque (in ft-lbs this time: (T in ft-lbs) = (T in oz-in)/12/16) into account. So, HP=(RPM*Stall Torque)/5252. This applies in gearing since you can modify RPM and torque, but you keep the same horsepower.

For our example, we found the 9V RPM to be 81 and the torque to be 4056 oz-in. Plugging that into the HP formula, the HP at 9V=81(4056/12/16)/5252=0.326 HP. Wow, that's less than I expected. And probably less than what you'll be dealing with (in 12-lbers, typically motors are around 1HP each, I believe)! But that's okay, this is all imaginary. Let's see what happens when we plug in the 12V stats. at 12V, the HP=(108 RPM * 5408 oz-in/12/16)/5252, which equals 0.579 HP! That 3V difference accounted for about twice the horsepower!

This leads me to my next point: over-volting. As it should be coming clear now, over-volting is a great way to get more horsepower out of a motor: doubling the voltage quadruples the horsepower! But this has its drawbacks. First of all, this also greatly increases the heat which the motor creates. If this heat gets to be too much, it may damage the motor. Secondly, make sure your gearbox can handle it! It may strip some gears. Thirdly, it greatly increases the amperage. This increases the torque, but also causes higher drain with your batteries. And fourthly, it causes a tendency to make the motors spark when energized, which admittedly has a coolness factor, but also can cause radio interference (to yourself, not others). Just make sure you know what you're doing. If you're using a common motor, ask around for recommendations on the max voltage to overvolt a motor. Don't overdo it!

One last thing: calculating the exact RPM at a given torque (this is again an ideal calculation, but it's more accurate than using no load RPM, and will get fairly close to the true value). This is an equation I've derived using some assumptions:
1) Kv*V=no load RPM
2) V/R=stall amperage
3) stall amperage*Kt=stall torque
4) Kt*Kv=1352
5) At Torque=0, RPM=No load speed
6) At Torque=Stall Torque, RPM=0
7) RPM and Torque vary at a constant rate.

Knowing all of this, I can set up a graph: Torque (the independent variable, what you plug in) is x-axis, RPM (the dependent variable, the output) is y-axis. The graph will be a line connecting the point (0, no load) and the point (stall torque, 0), which will form a triangle with the right angle being at the origin.

From there, I need an equation for torque. Y=MX+B. M is the slope, the rate of change, and is equal to (y2-y1)/(x2-x1). So, M=(-no load RPM)/(stall torque). Or, better, M=((-Kv)(V))/((V)(Kt)/(R))). The Vs cancel, leaving M=(-Kv*R/Kt), making this graph applicable to the motor itself. B is the y-intercept, or what y equals when x=0. In this case, B is equal to the no load speed, or Kv*V. So, back to our equation: y=mx+b. RPM=((-Kv*V)/(Kt*V)/R)T + Kv*V, where T=torque required to move the bot (in oz-in). The V's cancel, leaving the equation as RPM=(Kv*V)-((Kv*R)/Kt)*T. Factor out the Kv's to get:

RPM=Kv*(V-((R*T)/Kt))

Let's test it. With our example motor, we ran the motor at 9V. R=1/3, Kv=9, Kt=150.2. According to this equation, the RPM at no load (T=0): Kv*(V-0), RPM=Kv*V, which by definition is right. At stall, plug in T=4056: RPM=Kv*(9-(1352/150.2), RPM=Kv*(9-9), RPM=0. That by definition is right, too! But what if you want some real-life RPM? Suppose the bot needs to have 1000 oz-in of torque to turn its wheels. What would its RPM be? Plugging things in:

RPM=9*(9-((1000/3)/150.2)). RPM=61 .03. Haha! That took way too long. Don't follow what I did? No problem; just copy down that equation. Where it came from isn't as important as its use, although if you do follow my above math, it may help you understand the other equations better.. Or maybe it won't.

So, all the equations again!

Kv=(no load RPM)/V
No load RPM=Kv*V
V=(no load RPM)/Kv
Kt=(torque (oz-in))/(current)
Torque (oz-in)=Kt*(current)
Current=(torque (oz-in)/Kt
V=(stall current)*(motor resistance)
Stall current=V/(motor resistance)
Motor resistance=V/(stall current)
Kt*Kv=1352
Kt=1352/Kv
Kv=1352/Kt
HP=((stall torque (ft-lbs))*(no load RPM))/5252
No load RPM=(5252*(HP))/(stall torque (ft-lbs)
Stall torque (ft-lbs)=(5252*(HP))/(no load RPM)
5252=((stall torque (ft-lbs))*(no load RPM))/HP
RPM=Kv*(V-((R*T)/Kt))

If you substitute the above equations into other equations, you can come up with even more: For example, HP=((nls)*(stall torque (ft-lbs))/5252. You can rewrite this as HP=(Kv*V*Kt*V)/5252*R*12*16 (the latter two because Kt deals in oz-in and HP in ft-lbs, so you have to switch units. 1/(12*16) ft-lbs=1 oz-in.). Combining the numbers and V's, HP=(Kv*Kt*V^2)/1008384R (v^2=v*v in case some of you don't recognize this). A little complicated! But there's still more to substitute! Kt*Kv=1352, so HP=(1352*V^2)/1008384R, and simplifying it further gets you:

HP=(13*V^2)/(9696*R),
R=(13*V^2)/(9696*HP), and
V=the square root of ((9696*R*HP)/13)!

Handy equations to know, and you can graph the HP curve with it.

Last notes:
The following symbols that I've told you about are constants for a particular motor-
Kv
Kt
R (stays constant in the equations I've given you, at least)
(With the ks, remember, "k stands for 'konstant'"!)

The following symbols vary:
V (usually independent, or a target voltage to shoot for)
I
HP
Stall I
Stall torque
Torque necessary (independent in this case, usually, unless you're say solving for how much torque it would take to reach a dangerous current draw)
RPM
No load RPM

Make sure your symbols match up! When combining Voltage and Stall Current, for example, make sure the stall current you enter matches with the voltage! It can sometimes be easy to plug in the stall current at 12V in and 9V in for voltage by accident.

Past that, that is advanced motor math! If you understood that, great! That's probably the most complicated thing you'll have to learn in these basics web pages. If not, try rereading it. If that doesn't work, try taking a long break from this and rereading it later. You'll figure it out soon enough! Although if you have questions about a specific aspect, e-mail me about it, so I know I need to improve on that area!

Practice! (Chart of Popular Motors' Constants, Find their RPMs, stall current, torque, HP, etc)


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