Power from two sources, electrically assisted bicycle

[Matt atkinson]

So I'm just doing a project where i have a electrically assisted bicycle and I am struggling working out the power when the bike is going uphill.

If i have a certain force provided by the rider, and a force provided by an electrical motor; can I do Power=F_{motor}* v_{motor}+F_{rider}*v_{rider}

where the v_{motor} and v_{rider} are the speeds that the given force will generate, or would it be P=F_{total}*v_{total}.

sorry if this is a basic question I've just been struggling and needed some reassurance.

 

[BvU]

Both ! Does that offer some reassurance ?

There is a small snag: because of wind resistance, power is not a linear function of $v$ (the force to go twice as fast ($2v$) is more than twice the force for $v$).
So if you on your own go 15 km/h and the motor would give you 10 km/h, the two of you together won't achieve 25 km/h, but for example 20 km/h. But the relative contributions would still be 15/25 (60%) from you and 10/25 (40%) from the motor.

 

[Matt atkinson]

Oh that does, I didn't think of it like that!

So if we are traveling at a set speed (constant) say 40kmph for both the motor + rider contributions it would be better to work the power out using the second method i mentioned by working out the sum of the forces acting on the bike?

 

[BvU]

Yes. The $v_{\rm rider}$ and $v_{\rm motor}$ aren't very meaningful. $F_{\rm total} = F_{\rm rider} + F_{\rm motor}$ is much more sensible.

 

[jbriggs444]

Yes. The $v_{\rm rider}$ and $v_{\rm motor}$ aren't very meaningful. $F_{\rm total} = F_{\rm rider} + F_{\rm motor}$ is much more sensible.

Well, it might be more sensible if one had a definition for the three terms in that formula. No such definition is evident for any of the terms here.

 

[BvU]

I plead guilty to the well-meant reproach in post #5. $F_{\rm total}$ is easily defined: the force needed to maintain the speed $v$. It can be measured, e.g. with a spring balance. And then the other two have to be loosely defined as the (average) fractions times $F_{\rm total}$. Perhaps $F_{\rm motor}$ can be recovered from the specifications... (or from the time it takes the battery to go empty, an efficiency, etc. etc.). Even exact is relative .

 

[jbriggs444]

I'll buy that. Though this model has limitations. It seems to assume that one has a single-speed bicycle, a constant torque motor and a rider who will exert a constant average force on the pedals throughout a relevant range of speeds.

 

[russ_watters]

I'll buy that. Though this model has limitations. It seems to assume that one has a single-speed bicycle, a constant torque motor and a rider who will exert a constant average force on the pedals throughout a relevant range of speeds.

I don't see how it makes such assumptions. I only see force, not torque or rpm, so the drivetrain issues just seem to not be addressed. It might just be that the OP hasn't gotten that far yet.

If I were designing a motor assisted bike, I might put the motor on the input side of the drive so it can take advantage of the gears along with the rider.

 

[BvU]

Matt, I've seen what other threads you initiated and I'm wondering about this project. I would almost think one of your kids hijacked your account, but there must be a better explanation ?

 

[Matt atkinson]

Well, is might be more sensible if one had a definition for the three terms in that formula. No such definition is evident for any of the terms here.

I do have equations for those terms, I was just curious which way was best to approach given my situation where I'm trying to stay at a constant cycle speed.

Yes. The $v_{\rm rider}$ and $v_{\rm motor}$ aren't very meaningful. $F_{\rm total} = F_{\rm rider} + F_{\rm motor}$ is much more sensible.

Thankyou, it works out better this way!

I don't see how it makes such assumptions. I only see force, not torque or rpm, so the drivetrain issues just seem to not be addressed. It might just be that the OP hasn't gotten that far yet.

If I were designing a motor assisted bike, I might put the motor on the input side of the drive so it can take advantage of the gears along with the rider.

Yeah I haven't got that far yet, I've just been roughing out some physics first thanks for the suggestion.

Matt, I've seen what other threads you initiated and I'm wondering about this project. I would almost think one of your kids hijacked your account, but there must be a better explanation ?

Matt:
Yes my friend borrowed my account, they wanted to ask for a second opinion. Maybe i didn't explain it well enough thanks all! I need to up my skills as a tutor haha

 

[jbriggs444]

I don't see how it makes such assumptions. I only see force, not torque or rpm, so the drivetrain issues just seem to not be addressed. It might just be that the OP hasn't gotten that far yet.

You cannot add force determined by a spring scale attached to a bike at one speed and force determined by a spring scale attached to a bike at another speed and get a number that is meaningful for the purpose at hand if the speeds are different without such an assumption.