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- Joined
- Dec 13, 2012
- Messages
- 2
Hello,
I'm questioning the use of some formulas (the ones used to calculate energy and power on flat and climbing) in the event of downhill slowdown (not complete stop). I'm not sure about the formulas being adequate for what I use them to, so your help is greatly appreciated.
The idea is to compute how much energy can be harvested from going downhill at a reduced speed vs coasting on a heavy touring bike, and the corresponding power requirements.
It is NOT about the motor efficiency at given RPM or anything electrical. We only focus on mechanical energy. Electrical losses will be computed later.
I compute the energy and power from first calculating force vectors. Then we just have to multiply by the speed to get the power, and the distance to get the work (energy). I use SI and metric units. The system is at equilibrium, constant speed. We won't study the kinetic force of acceleration and stopping (regen braking) as it's not we're focusing on.
At constant speed, the sum of the rolling resistance + air drag + slope force equals zero. Fr+Fa+Fs=0
- Rolling resistance
Fr=Crr.m.g.cos(a)
With Crr the rolling resistance coefficient, g (gravity acceleration) of 9,81 m/s^2. For any reasonable incline (<20%), cos(a) -> 1 so if you feel lazy, get rid of that term (I'll plug it in a spreadsheet so I don't care). Crr for touring bike is around 0,01. I'll use a mass of 120kg.
- Air drag
Fa=(1/2).rho.Cda.(v+vw)^2
With rho density of air of 1,22kg/m^3; Cda the combination of S.Cx of 0,4 for a touring bike, v the speed of the bicycle, and vw the speed of the wind in the axis of motion.
- Slope force
Fs=m.g.sin(a)
Gravity my friends. m the mass (still 120kg), g still 9,81 m/s^2, and a the angle of the incline in ° (or rad). You'll get it from the incline in % by arctan(angle in %). (if you forget this step, the result will be less precise, but in the same ballpark anyway).
I want to climb a hill
So let's imagine a 10 km long, 10% incline, that I want to climb at 10 km/h with no headwind.
Fr=12N, Fa=1,9N, Fs=117N (gasp) so Fresist=131N.
That gives a consumed energy of 1310 kJ and a constant power of 363W
Now I'm on top, exhausted and I come back home.
It's still 10 km long, 10% incline (but this time is downhill). Let's say I go downhill 10 km/h too (for computing sake, no way in real life).
Fr=12N, Fa=1,9N, Fs=-117N. Same thing but with the slope the other way. Fresist=-104N
The energy I'll have to dissipate in the brakes (or harvest via regen) is 1040 kJ at a constant power of 288W given I don't pedal.
The equation allows me to know which speed I'd go without braking (or regen), that is when all the forces cancel out. It gives about 75 km/h
Which sounds coherent given the load and the slope.
It also tells me that if I wanted to go dowhnill at 40 km/h, id' be able to harvest 753 kJ but would need to dissipate or absorb 837W. Less energy because going faster makes wind rob me energy, but more power as I harvest this energy in less time.
If I check with the gravitational potential energy, Egp=m.g.(h2-h1). A 10% slope 10km long has a 1km elevation, m and g still the same, I get 1177 kJ, slightly more than computed at 10km/h which is logical: at 10 km/h I start losing energy by rolling resistance and air drag anyway. So it seems correct.
If I add the cinetic energy, (1/2).m.v^2, I get at 10 km/h I get some 463 Joules more, and at 40 km/h some 7,4 kJ more.
The question is: do these equations apply for regenerating? Are the energy and power computations correct? I would think that, yes, but the power seems quite abrupt to me.
Could you confirm or correct my assumptions?
Thank you,
Nicolas
I'm questioning the use of some formulas (the ones used to calculate energy and power on flat and climbing) in the event of downhill slowdown (not complete stop). I'm not sure about the formulas being adequate for what I use them to, so your help is greatly appreciated.
The idea is to compute how much energy can be harvested from going downhill at a reduced speed vs coasting on a heavy touring bike, and the corresponding power requirements.
It is NOT about the motor efficiency at given RPM or anything electrical. We only focus on mechanical energy. Electrical losses will be computed later.
I compute the energy and power from first calculating force vectors. Then we just have to multiply by the speed to get the power, and the distance to get the work (energy). I use SI and metric units. The system is at equilibrium, constant speed. We won't study the kinetic force of acceleration and stopping (regen braking) as it's not we're focusing on.
At constant speed, the sum of the rolling resistance + air drag + slope force equals zero. Fr+Fa+Fs=0
- Rolling resistance
Fr=Crr.m.g.cos(a)
With Crr the rolling resistance coefficient, g (gravity acceleration) of 9,81 m/s^2. For any reasonable incline (<20%), cos(a) -> 1 so if you feel lazy, get rid of that term (I'll plug it in a spreadsheet so I don't care). Crr for touring bike is around 0,01. I'll use a mass of 120kg.
- Air drag
Fa=(1/2).rho.Cda.(v+vw)^2
With rho density of air of 1,22kg/m^3; Cda the combination of S.Cx of 0,4 for a touring bike, v the speed of the bicycle, and vw the speed of the wind in the axis of motion.
- Slope force
Fs=m.g.sin(a)
Gravity my friends. m the mass (still 120kg), g still 9,81 m/s^2, and a the angle of the incline in ° (or rad). You'll get it from the incline in % by arctan(angle in %). (if you forget this step, the result will be less precise, but in the same ballpark anyway).
I want to climb a hill
So let's imagine a 10 km long, 10% incline, that I want to climb at 10 km/h with no headwind.
Fr=12N, Fa=1,9N, Fs=117N (gasp) so Fresist=131N.
That gives a consumed energy of 1310 kJ and a constant power of 363W
Now I'm on top, exhausted and I come back home.
It's still 10 km long, 10% incline (but this time is downhill). Let's say I go downhill 10 km/h too (for computing sake, no way in real life).
Fr=12N, Fa=1,9N, Fs=-117N. Same thing but with the slope the other way. Fresist=-104N
The energy I'll have to dissipate in the brakes (or harvest via regen) is 1040 kJ at a constant power of 288W given I don't pedal.
The equation allows me to know which speed I'd go without braking (or regen), that is when all the forces cancel out. It gives about 75 km/h
Which sounds coherent given the load and the slope.
It also tells me that if I wanted to go dowhnill at 40 km/h, id' be able to harvest 753 kJ but would need to dissipate or absorb 837W. Less energy because going faster makes wind rob me energy, but more power as I harvest this energy in less time.
If I check with the gravitational potential energy, Egp=m.g.(h2-h1). A 10% slope 10km long has a 1km elevation, m and g still the same, I get 1177 kJ, slightly more than computed at 10km/h which is logical: at 10 km/h I start losing energy by rolling resistance and air drag anyway. So it seems correct.
If I add the cinetic energy, (1/2).m.v^2, I get at 10 km/h I get some 463 Joules more, and at 40 km/h some 7,4 kJ more.
The question is: do these equations apply for regenerating? Are the energy and power computations correct? I would think that, yes, but the power seems quite abrupt to me.
Could you confirm or correct my assumptions?
Thank you,
Nicolas