My1020 Unite Motor

Lessss

1 MW
Joined
Nov 18, 2006
Messages
2,345
Location
Saint John N.B. Canada, Sol 3
I've a 900w motor from unite via defscooter.

I figure I'd go down to a couple of the local motor shops to see about mounting it and getting a sprocket for my chain type. No luck there and an icy shoulder to boot.

Anyway. I need some ideas on how to mount it and where I can find a sprocket for it for a typical bicycle chain.

mud2db9.png
 
Lesss, I can't help much with what your working on, but here are a few links that may lead in the right direction.

http://www.electricscooterparts.com/sprockets.html

http://www.monsterscooterparts.com/sprockets1.html

http://www.scooterparts4less.com/all_electric_scooter_parts.htm
 
how is this now?
with a temp sensor and an amp limit more v is Ok, but i'm not sure how high & how long.
 
I would make the gearing shorter then it should handle the step up in voltage without any problems. The torque load at the motor would go down so it should pull less current and thus less heat. Gear for the same top speed and it would run ever so slightly hotter than before depending on the increase in no load curren(which goes to increased windage,core loss, brush loss,etc), your only gain would be increased acceleration. Gear for more speed and heat will go up with the square of your increase in speed.
Joe
 
Joe
if fechter's current limiter was used, and set right, wouldn't speed w/ tailwind go way up?
say it is geared for x mph at 36v FOR ALL CONDITIONS,
then under ideal contitons and more v, it goes much faster and also if u lean & tuck.
 
Yeah it would because the current limiter is in effect limiting the voltage to maintain a certain current(torque). If you catch a tailwind or a downhill the torque required to hold that speed goes down and your speed goes up, at the same time if you go up a hill your speed drops. With a system like that your top speed will be damn near impossible to determine as it depends on your environment at that moment.
Joe
 
Joe
does it become more efficient at higher v?
 
Maybe yes maybe no. There are two components to look at when your looking at losses, first you have to look at loss then look at input power to get a percentage. Resistive losses will go down, usually thats the majority of your loss, you can see this on a motor curve no load losses dominate on the high speed low load side of the curve resistive losses dominate on the low speed high load side of the curve, when the two are equal your at the efficiency peak.
If you overvolt and keep the same top speed through gearing your effective load stays the same but your torque load drops. Now less torque means less current and less resistive loss, thats good it puts you closer to the efficiency peak. But theres the other side, spinning the motor faster will raise your no load current draw, this is the current required just to make the motor spin it goes up with speed, and it is a constant loss. Whether or not trading off resistive losses for a no load loss is dependent on your motor and your setup. Normally i would say that it is more efficient. however overdoing has other much more serious problems than just efficiency, such as the motor surviving the centrifugal effect, excessive brush wear, etc.
For a formula you have (noloadcurrent*input voltage+(current-noloadcurrent)^2*armature resistance)/input power
Joe
 
I should do a no load amp test next.
36v(i think it was 2-2.5 new with rear wheel up)
43-44v ?
48v?

I'm curious as to the increased loss. (Anyone care to guess?)
Good points! thanks!
 
slight disc brake drag.......
DRUM ROLL.......
36v.....2.8 amps

48v.....2.8 amps
does this mean go 60v?
 
Well staying the same seems impossible, but its obvious the current didnt go up that much. I would go for it, be sure to change the gearing to compensate for the extra rpm's, then report back and let us know how it worked out.
joe
 
OK DID 60v used turbo bypass.
3.6 amps!
Howz that?
 
Its actually surprising, but at the same time makes it more likely to work out.
In hindsight your efficiency peak will end up being at a higher load condition, if at full speed your motor is already heavily loaded this could help greatly, if not then you might actually lose efficiency by ending up on the other side of the efficiency peak.
If you know your current draw at load and the armature resistance it can be calculated
Joe
 
Honk from the VforV had a superb post on this. Motor efficiency may actually go really badly down in terms of % as voltage and current is increased. You need to find the internal resistance of the motor by applying a 1V DC power source to motor and measuring minimum volts at the motor. This will measure resistance of brushes, commutator and windings. If I recall correctly (and I probably don't), 0.75V at the motor = .75 ohms resistance and so on. That could be wrong, don't suppose anyone has that thread archived?

He even knocked up a simple program to calculate motor efficiency at various current/voltage combos. In fact, I've still got a copy and attached it. It's no good unless you know your motors resistance, but fun to play with.
 

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Actually i was just looking at the equation and realized that the term for the no load loss was just multiplied by the voltage so loss is proportional to no load current as well as voltage. This means motor heating will always be significantly greater. What you do gain is on the other term because the load current goes down and this term is squared.
The total loss under low load is greater, the rate of rise in loss is lowered. Add to this the efficiency peak will now be pushed into a higher power region of the motor. The loss alone doesnt give you an efficiency rating it has to be calculated with the input power, wether or not it gives you a greater peak efficiency needs to be calculated. Also if you have enough data and calculate this efficiency peak to coincide with your full speed load im sure there could be a reasonable gain in overall efficiency.
Joe
 
Leeps said:
Its actually surprising, but at the same time makes it more likely to work out.
In hindsight your efficiency peak will end up being at a higher load condition, if at full speed your motor is already heavily loaded this could help greatly,......Joe
500w 36v my1020,
at 36v 10 amps 17mph cruise, light load,

at 43v 20mph cruise, 14-15 amps, higher load
this is my favorite ~625w input ~500w output,

at 48v 23mph ~20 amps, thats 960w input, 768w output if at 80% eff.
Could eff go up by 5%? to 85%? then output is 816w. Or maybe it is only at 65%? 624w out?, goes up to what?
700w?

at 60v 28mph i'll guess 30 amps, 1800w input, this good for short bursts only, i can't imagine a 500w 1020 not burning up within about 10 minutes.

the 1200w 48v version is the same size and costs ~$100. that should work great at 60v. the mx500 is geared down as a dirt bike, so imo the extra voltage gives "gears" to a 1 speed. even 72v is only 34-35 mph. But with my mbike being un-aero, that would be short bursts only.
 
lemmiwinks said:
Honk from the VforV had a superb post on this. Motor efficiency may actually go really badly down in terms of % as voltage and current is increased. You need to find the internal resistance of the motor by applying a 1V DC power source to motor and measuring minimum volts at the motor. This will measure resistance of brushes, commutator and windings. If I recall correctly (and I probably don't), 0.75V at the motor = .75 ohms resistance and so on. That could be wrong, don't suppose anyone has that thread archived?

He even knocked up a simple program to calculate motor efficiency at various current/voltage combos. In fact, I've still got a copy and attached it. It's no good unless you know your motors resistance, but fun to play with.

What ever happened to Honk?
We could use him around here. I'd like to see how his axial motor is coming along.

Anybody got an e-mail address for him?
 
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