Watts per pound per MPH

[Fishmasterdan]

I have been searching and have not really came up with, or found what I am after.

I am looking for any data or formula for what it takes in watts to figure out per pound, for a given speed.

So based on a flat surface. I currently know that 1800 watts and 360 lbs yeild 30mph. What my bike does.
8% grade same 1800 watts yeild's about 18 mph. No pedaling.
Geared puma motor on a mountain bike.
If I was AUW of 200 lbs does my speed increase and watts go down???

I am not looking to caculate in wind or efficencies. I am looking for a general idea.
So how many watts does your system pull for a given speed. I am looking generally at the geared motors.

I picked up an eagle tree data logger to I can get some real data. It would be nice to tell someone they need to be able to pull XXXX amount of watts to do what they wanted, so they could narrow there search for motors down.

[TylerDurden]

Weight is not as significant as air drag. (I got schooled by some ES members on that, a while back. )

Weight effects are linear, whereas air drag effects are parabolic.

Weight is significant on hills, accelleration and rolling-resistance of the tires.

WeightvGrade1.jpg (34.64 KiB) Viewed 580 times

slopeVmph1b.gif (31.98 KiB) Viewed 580 times

[Doctorbass]

Here is mine based on true results obtained by an average result of many E-S member stat:

But it not take account of pound..

One represent power at the wheel and the other represent power draw on the battery... (with a 80% efficient system)

Doc

[Jeremy Harris]

As a general rule, air resistance dominates above a few mph, as has already been mentioned. Air resistance is proportional to the square of air speed, so power required to overcome that resistance is proportional to the cube of air speed.

For example, if it take 100 watts to do 10mph on a particular bike, then to estimate the power needed to do 20mph ( 2 x the speed) multiply the power for 10mph by 2^3 (2x2x2=8) which gives 800 watts. If you know the power needed for a particular speed, you can make a pretty good estimate of the power needed for any other speed using this simple formula.

Jeremy

[Russell]

As a general rule, air resistance dominates above a few mph, as has already been mentioned. Air resistance is proportional to the square of air speed, so power required to overcome that resistance is proportional to the cube of air speed.

For example, if it take 100 watts to do 10mph on a particular bike, then to estimate the power needed to do 20mph ( 2 x the speed) multiply the power for 10mph by 2^3 (2x2x2=8) which gives 800 watts. If you know the power needed for a particular speed, you can make a pretty good estimate of the power needed for any other speed using this simple formula.

Jeremy

Jeremy,

There are 3 drag components a cyclist must overcome; mechanical losses which are generally a few percent and relatively fixed, rolling resistance which increases linearly with speed and of course aerodynamic drag which increases exponetially. In your example of 10mph aero drag is actually quite a bit less than rolling resistance. At about 15mph wind resistance and rolling resistance are roughly equal. At 20mph wind drag is about double rolling resistance. Of course rider position, tires, and weight all affect the proportions for the various drag components but generally it takes from under 200W for a light rider in a full crouch on a light bike with high pressure tires to perhaps 300W for a heavy rider in an upright position on a heavy bike with knobbies to maintain 20mph.

On an electric bike cogging torque is an additional force to overcome. The power drawn from the battery is of course more than the figures above due to inefficiencies of the motor and controller which at cruise may be around 25%.

My old favorite bike calculator is gone it seems but this one works pretty well too:

http://www.mne.psu.edu/lamancusa/ProdDiss/Bicycle/bikecalc1.htm

-R

[Fishmasterdan]

Wow thats some great info exactly what I was looking for.

Now I am going to try and get some data about heat.

For example My puma motor ROR(rate of rise) in temperature is 20f at 1800 watts per 5 minutes at 55f. Crude but close measurement with my temp gun.

So I could reasonably say I can go up an 8% grade at 1800 watts with at 350 lbs at a 20f ROR temperature without risk of heat damage.
So to be safe I will shoot for 180f as the cut off point, I can easily go up an 8% grade for 30 minutes would put me at 175f.


I will plan for 20 mph, so essential I could go up a 10 mile hill at full throttle with my puma set up without worry of overheating. With mild pedaling.

Hows that theory sound??? Looks like I may have to try my monster hill nearby next.

I want to caculate out my bafang also and see what is realistic for that little thing without overheating it.

[Drunkskunk]

ROR isn't lineir. if the ambiant tempature is 80F, and your motor starts at 80F with a 20f ROR for 1 minute, in 2 minutes, there will be a 40 degree diffrence in temp, and a significant cooling effect. eventualy, depending on the motor's ability to transfer heat, the temp will stop rising, or slow significantly.

I can't get my motor over 150F or so. at that point, on an 80F day, the motor is in an enviroment 70F cooler, and the cooling effect is similar as having a motor at 80F, but on a 10F day

[Jeremy Harris]

Thanks for the lecture, Russell, much appreciated......

Rolling resistance is very variable, and depends on bike type, tyre tread pattern, tyre width and aspect ratio, tyre air pressure, wheel diameter, bearing drag, road surface, temperature and humidity. However, as I quite correctly pointed out, for any given bike, air drag tends to dominate at higher speeds, above around 10mph or so, and if all that is wanted is an estimate of power needed to achieve a faster speed, then simplifying the sums to just the cube law relationship between speed and power needed to overcome air resistance works well enough for most purposes. At the end of the day, most people are simply looking to size power systems, so are looking for motor, controller and battery capacity needed to achieve given performance levels on a particular bike.

Of course, if one wishes to undertake a full analysis to determine power needed for a specific bike, under specific conditions, to an accuracy of 1 or 2%, then it's worth looking at all the various drag factors in depth. However, this is a fairly pointless exercise, given the lack of hard data for some of the critical rolling resistance factors, if all you need is a power estimate.

Jeremy

[Russell]

Jeremy,

I don't disagree with the statement that aero drag is the predominant component as speed increases though I would place the crossover point a bit higher at around 15 mph where aero and rolling resistance are roughly equal, again depending on the bike, rider and equipment. I did take exception to your example which ignored the other components at 10mph which made your calculation of 800W at 20mph wildly inflated. Bike power calculators aren’t the end all by any means but they do allow a person to manipulate the variables to see how much each item affects power requirements. If you use the calculator you’ll find the aero drag at 10mph at the most is going to be less than 20W.

-R