I’m working on my senior design project and have gotten motors out of a toy truck that I need to determine the specs for.
I already know I can use a tachometer or strobe light for getting the rpm of a motor. However, how do I determine the max voltage that the motor can take so that I know that I’m looking at the correct rpm and overall voltage requirements of each motor? Also, how would you physically set this up using multimeters and a voltage source? (I’m assuming that I’d have to use a voltage divider if the voltage source is a constant voltage. i.e., 12V.) Can I use one of the charged batteries for a FIRST robot to provide the power source or am I looking at a dangerous situation?
Can I use one of the charged batteries for a FIRST robot to provide the power source or am I looking at a dangerous situation?
No do not use a twelve volt battery as a power sorce for a toy car motor. I doubt that your toy car motor is designed to use twelve volts. There are a bunch of solutions that I can think of for your problem:
You should be able to dervie the voltage constant/voltage ratings using these methods:
1)Find the toy car and find out what voltage batteries it used.
2)If you have acess to a power supply apply two differnt voltages to the motor and then find rpms’s. Dervie data using voltage equations.
3)If you have acess to variable drill press clamp the motor’s shaft to the drill press and spin it at two differnt speeds. Find voltages at each speed. Derive data using motor equations. If you want to know what these equations are there are a bunch of equations from which you can dervie those neat little graphs that first gives us every year. I would recommend you pick up building sumo robots, building combat robots, or I believe building robot drivetrains might have the equations.
Theoretically, you can keep increasing the voltage until the motor sucumbs to material stress failures and literally flies apart. So there is not really a “max voltage” for the motor. Of course, at the same time you will be dumping increased current through the coils, which will be heating up. At some point the coils will either melt and turn into a pile of slag, or ignite and release all the magic smoke. Actually, the only thing in question is which happens first: explosion or immolation.
Rather than try to determine the maximum voltage that the motor can take (before self-destruction), you really want to characterize the motor at a specific voltage. Since you retrieved the motors from a toy truck, just determine the normal operating voltage provided by the truck power supply (most likely a set of batteries). You will need to determine two performance points - no load speed, and torque at stall. Run the motor at the operating voltage with no load on it and determine the speed (use a tachometer if you have one, or juct count really, really fast while watching the output shaft). Use a clamp meter around just one of the motor leads to determine the current draw at the same time. Record the speed point on a speed/torque plot, and the no load current on a speed/current plot.
Then do the same thing when the motor has a load applied to it that causes it to stall. One easy way to do this is to put a short arm (12" is ideal, or 1" if the motor is very small) on the motor output shaft, apply the current for a short period of time, and incrementally add small amounts of weight to the end of the arm until the motor can just hold the arm horizontally without moving. Note the total weight applied (including the weight of the arm itself!) and the distance from the output shaft, and convert to applied torque (in lb-ft, oz-in, N-m or any other convenient units like furlong-slugs). Record that point on your speed/torque plot, and connect the two points. Ditto for the current draw at stall.
You now have the basic speed/torque curve for the motor at one voltage, and a matching speed/current curve. When a different voltage is applied, these curves will move up or down on their axes, but the slopes of the curves will remain the same. With this data, you can determine the speed/torque combination at any point on the curve, and the curve for any voltage for the motor. You can also determine the total mechanical power for any point on the curve (MechPower[in watts] = Speed [rpm] / 60 * torque [in N-m] * 2*Pi). Or you can just plug the values in to the great motor calculation spreadsheet by Dr. Joe in the white papers section, which will calculate the curves for you.
There were three motors that were run off of the same battery pack. Two of them appear identical. Since the third was different, wouldn’t the normal operating voltage be dependent on the voltage division created by the electronics of the truck? If this is the case, can just looking at the resistors and tracing where the leads go on the board serve as the information for calculating the necessary voltage division?
That depends. While it is possible to measure the internal resistance of each component in the circuit, and then reconstruct the circuit design to determine the voltage drop across each component and motor, it will be a tedious job at best. You need to worry about the voltage drop across each motor, not just the fraction of the total voltage in each leg of the circuit. In some cases, depending on the exact components in the circuit, it may not be practically possible to determine this.
For the specific problem you are posing, I would probably forget all the detailed circuit reconstruction. Just divide the total voltage provided by the battery pack by 3, hook up one of the motors, turn it on for a short period of time, and watch it carefully to make sure it is not overheating or spinning so fast that it seems out of control. If everything seems OK, go ahead and connect it to the test rig and turn it on longer while collecting data. If the motor does start to get warm, drop the voltage by 20-25% and try again until you get into a comfortable operating range.
To elaborate on what Dave said above, you can’t really do these calculations based upon the DC resistance of the windings alone because they only behave in a purely resistive fashion at stall (i.e., when the motor shaft isn’t rotating). Because the windings have an inductive component, they are able to store energy in the surrounding magnetic field when they’re energized, giving it back in the form of Back EMF (Electromotive Force) when the field collapses. If you’re interested in learning more (which I would encourage), have a look at this website or just google on something like “DC Motor Model”
Indie,
Dave and Kevin have given you some good info so far. There is a way you can easily back into the calculation if you can partially disassemble the motor. The wire in the armature is designed to sustain full stall current. If you can measure the diameter of the wire, throw in a little fudge factor and then take a look at a wire size versus current handling table, you will get a rough idea of the design for the stall current maximum. Say, that from the above you find the wire in the motor is about 25% larger than a wire that would handle 10 amps. Making the assumption that stall current is then 10 amps, you need a variable voltage source connected to the motor through an ammeter and switch. Start out with a voltage that is lower than the battery in the toy, stall the motor shaft and momentarily supply a voltage to the motor while observing the current. If it is close to 10 amps then the battery voltage is most likely the design voltage for that motor. If it turns out to be significantly higher, then lower the voltage and try again. A simple variable voltage supply is a “D” cell battery holder and a couple of test leads. Each battery is 1.5 volts, so 8 in series equals 12 volts, 7=10.5v, 6=9v, etc. Once you know the design voltage and stall current you can determine the no-load speed at the design voltage. As to the other parameters you can really study the motor curves for robot motors and attempt to put enough mechanical resistance on the motor to change output speed while monitoring current. With this data in hand you can make some educated guesses about the motor specs without needing to use all of the sophisticated equipment that motor designers use to check their designs.
Now the hint, if the motors have any markings on them at all, most manufacturers will build into the part number the operating voltage range. The common ones for toys being 12, 10, 6 for lead acid batteries, 12, 9, 6, 3, 1.5 for alkaline batteries, 12, 7.2, 3.6, 2.4 for rechargeable NiCads. However, a toy manufacturer may not try to stay within the design specs of the motor. They may for example be perfectly happy to run a 9 volt motor at 6 volts, not worrying about full stall current or speed. It allows them to reduce the size of the wire feeding the motor and back down on the transmission design by using the tradeoffs of a lower supply.
I have the motor and a set up for the testing. However, I was given the following spec to be tested for:
9 oz/in
Is this the same as 9 oz-in? If not, what is the difference between the two numbers?
Unfortunately, the person working on this particular subsystem has not been able to explain any of the calculations he has made to arrive at the above requirement, only telling me that it is correct. What I do know and understand of the subsystem is that there is a “conveyor” belt that will be driven completely by the friction created between the belt and a roller. (i.e., No timing pulley.) When discussing the issue of the forces to be calculated, I was told by the person that the forces between the metal contacts of the roller “axle” (very small rod) and the side wall support were more important than that between the belt and the roller. Can someone explain why this would be true? I’ve been having a real hard time comprehending this since I always thought that the real driving force in these calculations would be the friction force, much like a tire against a road.
Indie,
Even I am a little confused. It sounds like you have the motor driving a simple two pulley belt drive where the belt is actually the conveyor belt. From what you said i gathered that the frictional forces at play on the pulley shafts and side load on the bearings are the significant frictional forces and I would agree with that. Remember that the conveyor belt moves because it has a high friction coefficient with respect to the surface of the pulley. Since one of the pulleys is driven by the motor, the pulley/belt friction is a small part of the overall load placed on the motor. The greater frictions are those produced in the pulley bearings by the tension of the conveyor belt and whatever load is actually placed on the conveyor, i.e. that material that is moved by the conveyor. The friction on simple sleeve bearings skyrockets as the belt tension is increased. It is for this reason that many belt systems use better bearings (roller or ball bearings) to minimize this friction. If one of the pulleys is on the motor shaft without the use of external bearings, expect high friction at the motor bearings perhaps resulting in motor failure. I saw at least one robot in competition using the chalupas with chain sprockets attached to the motor shaft. The resulting forces tore those motors apart in one day.
Your question on measurements 0z/in oz-in is correct, they are the same.
I downloaded Dr. Joe’s Motor Calc spreadsheet, but in looking at the data voltage I’m a bit confused. Is this supposed to be the voltage measured during the test, or the voltage it was supposed to be? For example, I was testing the motor at what was supposed to be 6V, but it was really 4.7V. Which number should be entered?
Indie,
I will defer to Joe to correct this but it appears that the data voltage is what you would measure at the terminals of the motor under test. Since Joe’s calculations are used for input electrical motor power the only correct calculation is motor terminal voltage vs. current. the fact that you were using a 6 volt battery and were reading much less is due to a discharged battery or a high current. Remember that there is some internal resistance in the battery that is electrically in series with the load. As a battery is discharged, this internal resistance goes up. More resistance limits the amount of current that can be delivered by the battery. Please remember that a battery that is not designed to supply the current needed by the motor will also show a reduction in voltage. I think it is important to note that the spreadsheet is to be used for the motors listed. I would contact Dr. Joe (PM) directly if he does not answer here just to be sure.
