0-15mph acceleration of 250W electric bicycle?

anotherkiwi

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I have never seen a pedelec with a throttle. I have seen an illegal Chinese e-bike (1 kW motor) with a throttle, once, never saw it again despite it being in a tower block in my GFs immediate vicinity.
 
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EddiePJ

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Leaving aside terrain and the like, I should have added that if I had to pick a hub drive or a crank drive bike to complete this challenge, then for me it would be hub drive every time.

That opinion is based purely on my experience of running both systems, that with a hub drive bike, the transition from zero to motion and increasing speed, is more fluid. With a hub drive bike, the gear changes always seem smoother with more precision. Having said that, the Pro Connect that I recently rode that had a Panasonic motor, was seemingly very smooth, quiet and nice to use.

If both bikes were fitted with single speed identical final drive gearing, I would have no idea which would be quickest, but I still think that I'd opt for the hub drive bike.
 

anotherkiwi

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I'm not sure Eddie. You may be right, but I feel that if I am in the right gear and assistance level at the lights there isn't much in it between mxus and GSM. When I was running the GSM unrestricted there were very many puzzled car drivers trying to understand why they had to try harder to keep up...
 
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D8ve

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Jan 30, 2013
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Six e bikes been in this house and none with throttle, although I do have one for the BBS just never fitted.
And I did live in Telford but left so must be fitter now.:D
 

Danidl

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Lets estimate:
The average torque of a 250w motor below 15 mph would be about 24NM. You can get that from the Ebike.ca simulator.

The radius of a 26" wheel is about 30cm, so the force at the tyre would be 24/.3 , which is 80N.

Acceleration is force divided by mass. Lets say a 75kg rider on a 25kg bike, so 100kg total. That means that the acceleration is 80/100 = 0.8 M/S/S

The first equation of motion says that final velocity = initial velocity +acceleration x time or V=U +AT. Turning that around, you get V-U=AT or T=(V-U)/A

U is 0, V is 15mph, which is 15x1500/ 60/60 M/S = 6.25 M/S

Therefore we have T = (6.25-0)/ 0.8. which is 7.8 seconds.
QED.
Alternative method
Mass of bike and person 100kg
Final speed 25km/ HR or 6.25 m/s. V ( not sure exactly how you got this number my calculation for 25km/HR is 6.9m/s but to maintain comparison I have used it)
Kinetic energy of bike 0.5mass x speed squared. 1952 joules
Energy input 250 watt or joules /s
Time 1952/250. 7.8seconds
Assuming 100% in motor output goes into forward motion no slipping , no wind, level ground
 
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Deleted member 4366

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At first, I was surprised that the two calculations gave the same result, but then I thought about it. My calculation was based on the average torque that an average 250w bike makes, using measured data. It seems to correlate very well with an average output power of 250w. That means that, although your bike will probably take as much as 600w or more from the battery under some conditions, it's average power consumption would be much less. That's when going for it with maximum power.

FYI I only estimated the conversion between MPH and M/S, which is why we have a difference.
 

Woosh

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In the first second, the bike is constrained by the maximum torque that can be achieved by the bike's specific motor at very low speed. The power that the system produces can be deduced from P = F * V.
F is equal to torque divided by the wheel radius.
P will rise with V until P becomes limited by the controller, then Danidl's method is the easiest to deduce the time it takes to reach the desired speed.
 

Danidl

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In the first second, the bike is constrained by the maximum torque that can be achieved by the bike's specific motor at very low speed. The power that the system produces can be deduced from P = F * V.
F is equal to torque divided by the wheel radius.
P will rise with V until P becomes limited by the controller, then Danidl's method is the easiest to deduce the time it takes to reach the desired speed.
Of course all of these methods are approximations. Neither of us have taken into account the energy held in the wheels. Assuming that the wheels including tyres are 3kg each this is an additional 6kg (above the static mass included in the previous calculation) of rotational kinetic energy . This is. 0.5 *6kg*6.25^2. 117.joules .
This is one case where the crank drive motor scores as it has less rotational mass compared with hub motors.
 

Woosh

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if you do that, you should also take into account air resistance, worth about 100W to 110W at 15mph.
 
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This is one case where the crank drive motor scores as it has less rotational mass compared with hub motors.
The crank motor has to keep speeding up and slowing down as you go through the gears, so you have to keep destroying its kinetic energy, then re-energising it. I can't see that as any advantage. All the enery going in to the hub-motor is conserved.

The crank-drive takes an immediate 5% to10% hit on its efficiency too because of the losses in the drive train.