Physics question

Baboonking

Pedelecer
Nov 4, 2006
147
6
Watford
I've got one of my powabyke motors in bits and was tipped to measure resistance of the brass brush holders, through which all current must pass to reach the brushes They both came in at a whopping 0.7 ohms. Was wondering what the formula was for working out how much energy is lost as heat in these parts. The battery voltage is 36 volts and the motor draws 20 amps peak. Perhaps an average of 8 amps during normal riding. Fortunatley its a nice easy modification to bypass them but I'm curious to know how much energy I would have lost.
 

Ian

Esteemed Pedelecer
Apr 1, 2007
1,333
0
Leicester LE4, UK.
Hi Baboonking

0.7 ohms seems very high, at 20A the voltage drop across each brush would be 20 x 0.7 = 14V (Ohms law, V=amps x ohms), ie 28 volts total leaving very little for the motor which is why a figure of 0.7 ohms is unrealistically high.

The square of the current times the resistance will give the power loss in watts, so at 20A each brush holder will dissipate 20x20x0.7 = 280W, again an unrealistically high power which would quickly heat the brush holders to melting point.

It is almost certain that most of the resistance you measured was at the contact point between the meter probes and the brush holder, brass is a very good conductor and normally something like that would have an almost immeasurably small resistance.

Ian
 

Grandad

Pedelecer
Mar 16, 2007
97
0
Devon
grandads.googlepages.com
I agree with Ian

Watts=I(current)squared R(resistance)
So for I=20A and R=0.7 ohms
W=400x0.7 =280Watts producing heat

And for I=8A and R=0.7 ohms
W=64x0.7 =44Watts producing heat

This seems to be a doubtfully excessive heat loss for each brush holder.

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As W=VxI @ 36 & 20 W=36x20 = 720W total power from the battery.

Or W=VxI @ 36 & 8 W= 36x8 = 288W total power from the battery.

As the Powerbyke seems to have a 200 watt power output DC brushed front motor this again doesn’t seem to add up.

I would guess that the max current the battery should supply would be I=W/V or I=200/36 = 5.55Amps

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So something appears to be wrong with your measurements. Try shorting your testmeter leads together to make sure that you get a zero reading.

On the other hand I may be wrong.:(

Hope this helps
Jeff
 

Ian

Esteemed Pedelecer
Apr 1, 2007
1,333
0
Leicester LE4, UK.
In reply to Grandad,

the 200W motor rating is it's continuous output rating, it short term peak will be much higher than that, a figure of about twice the continuous rating is typical.
A peak current of 20A on a 36V system sounds about right and probably equates to about 400W maximum output from the motor when electrical and mechanical losses are taken into consideration.

The nominal verses peak figures can be very confusing but are one reason why apparently similarly rated motors can perform very differently. When you open the throttle wide on a hill the motor is delivering it's peak power and it's nominal rating quoted in the sales brochure becomes largely irrelevant.

Ian.
 
Last edited:

flecc

Member
Oct 25, 2006
52,763
30,349
For reference the peak on that Powabyke motor is 700 watts, the highest of any British legal bike to my knowledge. The power "curve" is a steep spike.
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Last edited:

flecc

Member
Oct 25, 2006
52,763
30,349
These brush motors on our bikes all seem to have this steep spike of peak power, the other two, eZee Rider/Liv motor and the Heinzmann both 600 watts.
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Baboonking

Pedelecer
Nov 4, 2006
147
6
Watford
Gentlemen,

Thanks for the replys, your absolutley correct. The reading was way too high, so I went back and checked and couldn't get a resistance reading at all. Perhaps a normal digital multimeter isn't a very good way to measure resistance levels below 1 ohm anyway. For safe measure I may bypass the brush holders, though I should add I've no idea if there is an actual problem there at all.

I'll post some pictures of the motor in bits soon.