[Electric Boats] Re: gearing question....Myles

Thank you Myles....  gonna have to read this a couple of times to get all i can from it....  quickly though, if you set the voltage to a motor so that the motor wants to spin up to 1500 rpm but then apply a variable load so that the motor is only able to develop 1400 rpm (ever increasing load), doesn't the amperage draw of the motor then continue to rise until the motor destroys itself?

your basically saying the work done equals force applied times distance moved right?  (work done = torque * diameter*rpm)....  trying to recall old high school physics and thinking that what i was not thinking of correctly before and what you are attempting to explain to me is that "work done" is what i wish to maximize

also, in terms of cooling, doesn't it make sense to provide perhaps a fan system that is optimized to cool the motor rather than relying on what i understand to be usually quite clunky fan systems that are integral to motors such as the ME0913?  one can move quite a lot of air for the cost of very few watts

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From: Myles Twete <matwete@comcast.net>
To: electricboats@yahoogroups.com
Sent: Sunday, February 26, 2012 1:59 PM
Subject: RE: [Electric Boats] gearing question....

 

The magic of gearing is little more than the physics of trading torque for speed while maintaining power constant.

As an equation:

Power = Torque * Speed

                Power Out  = Power In – Power Loss

                Torque Out * Speed Out = Torque In * Speed In – Power Loss

Ignoring Power Loss for now:

                Torque Out * Speed Out = Torque In * Speed In

                Torque Out / Torque In = Speed In / Speed Out

So the more efficiently you can implement your gearing, the closer you come to the last equation above.

 

So say you need 100 torque units delivered to your prop at 1500 RPM.

And say that you have an electric motor rated at a maximum steady speed of 4500 RPM and continuous torque max of 30 torque units.

Is it sufficient to direct drive your prop with the motor?

While the motor can easily reach the target 1500RPM, in doing so it will be loaded down by 3.3x its continuous rated torque level.  This is also 3.3x its maximum current level assuming this to be a permanent magnet field motor.

The motor will not last long at that loading due to internal heating equal to continuous current of 3.3 times its max current level.

Now, if instead of direct drive you choose to put in place a 3x gearbox of some sort.

In this case, at 1500RPM prop speed the motor will be spinning at its rated continuous speed of 4500RPM---i.e. a 1/3 speed gain.

But what of the torque?  The torque will effectively be amplified by 3x by the gearbox.  In this case, the 100 torque units required at the prop will translate into 33 torque unit demand from the motor---slightly exceeding the motor's steady state limit.

What results?  The motor is able to reach the target speed, but still cannot maintain it due to the current slightly exceeding its limit.

At that point, it may be decided that it is acceptable to achieve only 1400RPM or so allowing the current to be below limits.

WE might think that using a 3.3:1 gearbox will achieve our needs better.

However, look at what happens: Motor hits its maximum 4500RPM, prop speed is only 1364RPM  and torque at the prop something maybe only 60 torque units.  At this speed ratio, the current ratio of motor MAX torque to prop torque is only 2x…

 

So maybe a 3.1:1 gearbox would be optimal…anyway, you get the picture.

Not considered in any of the above is the efficiency gain in running the motor at higher RPMs to achieve a given power output.  For that story to be understood, please look carefully at motor curves.  Even if your motor can "handle" the maximum current of your prop and can achieve much higher RPM than the prop needs, you may be accepting significant power losses in the motor as a cost.  For permanent magnet motors, torque is directly proportional to motor current.  SO if you accept that your motor will reach a top speed of only 50% of its rated steady top speed, then you are accepting that you will be running the motor with 2x the torque that it could be run at if it had a 2:1 gearbox.  And that 2x torque requires 2x current.  But here's the rub: POWER LOSS in the motor due to its internal resistance is proportional to CURRENT SQUARED!!!  So what appeared as a benign, so what, big deal 2x current increase really means 4x the power loss in the motor.  To put numbers to this, say your motor can deliver a decent 85% efficiency when appropriately 2:1 geared and delivering 3HP.  The internal power loss expected: 0.45HP, or 337watts.  Now, if instead, we coupled this motor direct and were able to achieve the same speed with the same load.  In this case, the motor speed would be ½ the earlier speed and so for the same power delivered, we expect 2x the motor current or 4x the power loss earlier---i.e. 1.8HP or 1348watts of heat loss in the motor!!!  You might burn this motor up quickly.

 

Bottom line is: Yes, there are losses in gearing, but they can pale in comparison to motor heat losses.  In the above example, the increased power loss of 1.35HP amounted to an overall efficiency to load of 63% compared to 87% with the gearing.  Since gearing options typically don't come anywhere close to 14% loss, the advantages of adding appropriate gearing are evident.

 

In case this helps-

 

-Myles Twete, Portland, Or.

 

 

From: electricboats@yahoogroups.com [mailto:electricboats@yahoogroups.com] On Behalf Of Michael Mccomb
Sent: Sunday, February 26, 2012 7:59 AM
To: electricboats@yahoogroups.com
Subject: [Electric Boats] gearing question....

 

 

does gearing an electric motor make any sense?   because electric motors generate the same torque at all revs doesn't gearing them do nothing but introduce a bit of power loss at the prop due to whatever friction is present in the reduction mechanism?   i guess you'd have to eliminate cooling via a motor incorporated fan as a concern...