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Simon G
01-17-2002, 06:59 PM
Ok, to start, I'm an aerospace engineer, not an ME. So gears are a new thing for me!

I'm trying to make a gear box for the chipuah (sp?) motors, but I'm banging my head against the wall, because it seems to me that I will surely shread gears with this monster of a motor.

According to my calculations, the tooth load on the pinion of the motor, at max 315 oz-in torque, tops out at 126.4 lbf. (19.688 in*lb of torque, 0.3115 pinion OD)

So at that force, according to my berg catalog, I should be shearing off the teeth of the 44T gear. That can't be right!

They give me a calculation of:

Wt (force transmitted) = (S*F*Y)/(DP)
where:
S is the max tooth stress (brass alloy 19ksi)
F is the face width (0.5" for the 44t gear)
Y is the lewis factor (0.399)
and DP is the diametral pitch, 36

Thus I get Wt of ~ 105 lbs which is < 126 lbs = BAD!

Am I being overly conservative???
What am I doing wrong???

Thanks,
Simon

Joe Johnson
01-17-2002, 07:28 PM
In my opinion the Lewis calculations are much more conservative than you need to be.

I typically put the tooth load on one tooth at the tip, find the distance to the smallest cross section then use a standard beam approximation to approximate the max. stress.

It is only an approximation. But it is seems to work tolerably well.

Joe J.

Lachuck894
01-17-2002, 09:02 PM
:confused:

For a non-engineer person, explain what was said above in layman's terms? :confused:

dense
01-17-2002, 09:36 PM
Would the torque still be realtively high if we reduce the rpm's of the Chiapuah to about half using PWM's and use the cluster gear for further reduction? (What is the exact ratio on the cluster gear, hehe, i keep on counting odd)

How much torque would that be in Newton meters or Foot pounds?

Joe Johnson
01-17-2002, 10:10 PM
http://www.uoregon.edu/~struct/courseware/461/461_lectures/461_lecture38/461_lecture38_pic2.gif

This is a big topic...

I found a lecture online that is not too bad:

Beam Stress Lecture with example problems (http://www.uoregon.edu/~struct/courseware/461/461_lectures/461_lecture38/461_lecture38.html)

http://www.uoregon.edu/~struct/courseware/461/461_lectures/461_lecture38/461_lecture38_pic3.gif

The bottom line is that you need to use the formula

S = Mc/I

S=stress
M is moment on the section of the beam in our case the height of the tooth times the stall torque of the motor (with a safety factor if you want -- I calculate the stress first and put in the safety factor afterwards)
c is 1/2 the thickness of the tooth at its weakest point near the base
I is the standard I (moment of inertia) for a rectangular cross section 1/12 b h^3 where b = width and h is the height (in our case the thickness of the tooth).

Solving you get:

Stress = 6 * Ftooth * ToothHeight / (ToothFaceWidth * ToothThicknessMin ^ 2)

Look on a table to see what material you are using and what stress it can take. I would not go above 30,000psi for unheated steel if I an help it (this is where I have my safety factor).

This is a very crude approximation but it is good enough for most of what I do in my daily job.

That is about as simple as I have time for right now.

Good luck.

Joe J.

jonnywalk
01-18-2002, 02:15 PM
If my understanding of the defination of torque is correct and i think it is then the output torqe of the motor is related to the voltage and no matter how you gear it you still get the same torque but by gearing you are essentially (and I say this for simplicity's sake) spreading the torque out over a larger time or concentrating it. The real advantage to gearing is you keep the motor turning so you don't stall it out. The same principle with a car, you have the same torque and horsepower in every gear but unless you car is really REALLY powerful you have to reduce it so you don't stall out until you get into higher into the power band. Hope this helps some.

jeffreym
01-18-2002, 08:54 PM
I don't know if this will help but I don't think the gear is brass. It seems to be made of a hardened steel. We had trouble drilling it and had to resort to ESD type of drilling.

iscrc2
01-19-2002, 02:09 AM
Simon, were you able to do the calculation using Joe's formula? How much of a safety factor do we have? If jeffreym is correct and the gear is harden steel then using 40ksi in your first formula you come out OK, right?

I'm confused about the sizing the gears as well. When I look in a gear book it seems that to carry the loads our systems are developing the books are recommending HUGE gears (like 4" pitch face, 1" shaft).

Is this because the books are for "production" levels and in fact when we run our 'bots we are pushing limits?

Matt Leese
01-19-2002, 08:12 AM
Originally posted by jonnywalk
If my understanding of the defination of torque is correct and i think it is then the output torqe of the motor is related to the voltage and no matter how you gear it you still get the same torque but by gearing you are essentially (and I say this for simplicity's sake) spreading the torque out over a larger time or concentrating it. The real advantage to gearing is you keep the motor turning so you don't stall it out. The same principle with a car, you have the same torque and horsepower in every gear but unless you car is really REALLY powerful you have to reduce it so you don't stall out until you get into higher into the power band. Hope this helps some.
No, you get very different torques by gearing it differently. Torque is basically an angular force. Gears are basically like levers; depending how they work (ie larger or smaller gear than the one before it) they either increase or decrease your torque. They also, inversely to torque, decrease or increase your speed. Now, as far as voltage fits in, voltage can change the amount of speed and torque. But usually when we do speed/torque calculations we assume a maximum voltage as that is where the motor is most efficient and gives the best results.

Matt

Lloyd Burns
01-19-2002, 09:27 AM
Originally posted by jonnywalk
If my understanding of the defination of torque is correct and i think it is then the output torqe of the motor is related to the voltage and no matter how you gear it you still get the same torque but by gearing you are essentially (and I say this for simplicity's sake) spreading the torque out over a larger time or concentrating it. The real advantage to gearing is you keep the motor turning so you don't stall it out. The same principle with a car, you have the same torque and horsepower in every gear but unless you car is really REALLY powerful you have to reduce it so you don't stall out until you get into higher into the power band. Hope this helps some.

In a car, which you mentioned, the [horse]power depends on the speed of the motor, and the torque is different function of motor speed. The power available at a given engine speed is constant through the gears, but the torque changes with the gears - you try never to start in third gear because, while it gives speed, it doesn't give enough torque to push the car from a stop. Once we've passed throught the torque curve a few times, we can get into top gear, and feel the wind.

Matt is correct: the product of the torque and the speed remain constant (power). If the second gear goes half as fast, the torque available doubles, less a bit for friction.

-=-=-=-

But, with electric motors, torque is related to current, and speed is related to voltage. And they are locked in a struggle with the internal 'resistance' of the motor.

The voltage applied to the terminals has to drive current into the armature, which if turning, generates a voltage to oppose the current coming in. Imagine a square, with a battery at the right side, + terminal on top, a resistor across the top, a generator with + terminal on top on the left, and the bottom of the square is a wire joining the - terminals of the battery and generator. The generator is the spinning armature of your motor; the resistor, the combined effect of armature resistance and the brushes.

The current that can enter the motor is given by Ohm's Law: current equals volts divided by resistance. The voltage here is the Applied voltage minus the Generated voltage. The torque is in proportion to the current, so at start, with the armature not spinning (just as when the armature is locked by a heavy load), the current is limited by resistance only, and the torque is tremendous.

As the torque causes the armature to spin, the voltage generated increases, causing the amout of current to drop, since the voltage difference the resistance feels is less. The torque drops off, the armature's acceleration decreases. If there is no load, the armature will speed up with decreasing acceleration until the current is all but cut off. At this free speed, the generated voltage is almost equal to the battery voltage. At high speed (free running), there is virtually no torque (just a little for friction).

When you apply a reasonable load, the motor will speed up to where the generated voltage limits the current in the resistance to the current which will generate exactly enough torque to overcome friction internally and drive the load at the speed. Increase the load (ram into the end wall, pull a goal) and the motor must slow down, to let the current increase the torque. Too much load, means no speed, and the rotor locks, and you get (in the Chiaphuas) 170 A, briefly, we hope.

So much for electric motors 101. It's a long course,:eek:

Ken Leung
01-19-2002, 12:23 PM
The most important about the electric motors we get from the kit is the speed torque curve. A lot of people don't understand how important this curve is, but you should really pay attention to it. Especially since woodie mentioned it a lot during kickoff.

What's so important about it is that it describe the behavior of a motor under load. Not what happen when you gear the motors or whatever other situation people came up with (which, by the way, was quite interesting to hear how people explain the speed-torque curve)...

So basically, the speed torque curve explains how a motor will react under load. The more load it face, the slower it goes, until the motor stall out. At no load, the motor is spinning at free speed, and under extremely amount of load, the motor is stalling and not spinning at all. And of course, as the load increase, the current it draw increase too.

So why is this important, well, it's because using this curve, you can figure out where are the most optimal places on the speed torque curve you want to me in. Either near the power peak, or the efficiency peak. And you do that by gearing down on whatever component you are using that motor on, and choose how much load it's going to reflect back to the motors. If you have a huge gear reduction, the motor will feel so little load that it might run near free speed... On the other hand, if you aren't gearing the motors at all, they will face all the load and stall if that amount is too big for them to take.

The amount of torque you can get out of is the range of torque from the speed torque curve, from zero to... well, not stall torque, because that's when you draw so much current that the circuit breaker will trip. Just look at the graph and figure out highest torque you can get before breaker start tripping, and multiply that with the gear ratio.

That's the range of torque your component will provide at different situation... Under load or no load. Say it's an arm, the arm will be pushing harder and harder within that range, before it start tripping breakers or start stalling the motor.

Kit Gerhart
01-20-2002, 09:31 AM
I'm sure most people on this thread are aware by now, but torque/rpm curves of DC electric motors are MUCH DIFFERENT from those of internal combustion engines. The basics of the motors, at constant voltage input are:

1) The maximum torque of the electric motors is at zero rpm (stalled)

2) The torque, and the current draw of the motor are inversely proportional to the RPM as controlled by load

3) The maximum power output of the motor occurs at about half the no-load rpm

While it has little to do with our robots, it is interesting to note that most electric vehicles such as golf carts, fork trucks, and pure electric cars do not use gear change. For those purposes, you can get adequate acceleration at low speed while using gearing selected mainly for desired 'high speed' operation.

With our robots, though, it is desirable to go as fast as possible, but also be able to push and pull very hard at low speed. Changing gears makes this possible without pulling lots of motor damaging and breaker tripping current from our motors.

Dreistul
01-23-2002, 11:35 PM
Originally posted by Joe Johnson

Solving you get:

Stress = 6 * Ftooth * ToothHeight / (ToothFaceWidth * ToothThicknessMin ^ 2)

I'm a little confused with this equation. What is FTooth? Is that supposed to be the force applied to the tooth, (in other words, FTooth=Torque/GearRadius?)

Also, by using this equation, does it mean that the speed at which the gear is turning does not have bearing upon the amount of stress applied to the gear?

Thanks for the help!

Chris Hibner
01-24-2002, 09:11 AM
Originally posted by Dreistul

I'm a little confused with this equation. What is FTooth? Is that supposed to be the force applied to the tooth, (in other words, FTooth=Torque/GearRadius?)

Also, by using this equation, does it mean that the speed at which the gear is turning does not have bearing upon the amount of stress applied to the gear?

Thanks for the help!

1) FTooth is the linear force at the tip of the gear tooth. So, you are correct that FTooth = Torque/GearRadius

2) This equation does imply that the speed of the turning gear has no bearing on the stress in the gear. Carl Barth (19th century) introduced the "velocity factor" for gears to take into account increased stress at high RPM. However, it is not a very exact calculation and I don't think it's worth worrying about for the short cycle times that our robots live through. Just use Joe's equation and give yourself a good safety factor and you'll be fine.

Joe Johnson
01-24-2002, 04:31 PM
There are a lot of simplifications to the formula, but this should get you in the ballpark.

The velocity factors that folks use are for hours and hours of continuous duty. We run for a few minutes at a time for a few minutes (in gear life terms) total.

The biggest safety factors are that I put the load at the tip of the tooth and I assume that only one tooth has the load. If you keep your pinions large enough (i.e. with 16 or more teeth, which I know is not possible with the Drill, Jideco and the Chiaphau motors), the contact ratios are large enough that for most of the travel of the gear tooth, more than one tooth is taking the load.

The biggest thing working against you is that no dynamic affects are included. This can be a very big deal.

In the end, we don't have time to really study the teeth to optimize them. Some years, my gearboxes are wearing out by the second regional, some years they are unworn after several post season competitions and hours of driver training.

When you get more experience, you will have better feel for the size you will need and where your approximations will get you in trouble. It also helps to get a few initials after your name ;-)

Joseph Johnson, Ph.D., P.E.