D

#### duivendyk

##### Guest

The math is as follows P= 0.193*T* N (trust me)

( the star denotes multiplication).The units are as follows:

P represents power in HP ( horsepower)

T represents torque in ft.lbs ('#)

N represents krpm ( kilorevs/minute,take note!!)

Power,that is energy delivery per unit of time is proportional to both torque and speed. For instance take the Robin EH 035 engine at 6000 rpm (N=6.0), the max torque is 1.6 ft.lbs,therefore P=0.193*6*1.6,or P=1.85 HP. At 7000 rpm we get T=1.3 and P=1.76 HP.

This is all well and good,but surprisingly small engine specs are not easy to come by,the Robin internet site is an exception (www.robinamerica.com),there is a wealth of excellent info to be found there.Information on the other engines appears to be lacking,does anyone have any info?

This torque business might seem a bit abstract but it is possible to relate it to tractive effort ,that is force where the rubber meets the ground:

T= moment arm* circumferential force, for instance a 26" wheel has a radius of about 1ft (31 cm),assume for instance we have 20 lbs road resistance ,it takes 20 ft.lbs torque..What is then the power required to do 20mph.? ,20mph corresponds to 270 rpm at the wheel (I will spare you the calculation,it is straightforward). So P= 0.193*20*0.270, or P= 1.04 HP. Assuming 6000 engine rpm ,the reduction ratio becomes 6000/270 =22 and the required engine torque 20/22 or 0.9 ft.lbs.This neglects drive train losses.So it probably takes at least 1.0 ft.lbs.

To the best of my knowledge the small 4 stoke engines have fairly flat torque characteristics peaking at 0.7/0.8 of max rpm.,2 strokes have a more pronounced power band,pretty much dying below 0.5 max rpm,but that may not be of much consequence.(The centrifugal clutch has not engaged).So knowing the Pmax it is possible to make reasonable guesses about the max engine torque.

It is of interest to take a look at the power capability of the NuVinci hub,its max torque spec is 95 ft.lbs. this means that at 10mph or 130 rpm with a 26" wheel,we get

P= 0.193*95* 0.13 ,P=2.4 HP max.This assumes no contribution from the cyclist.It also assumes a 1 to 1 hub transfer ratio.At maximum reduction 1/1.87 the road speed would become 5.4 mph.Some caveats apply,it is not 100% certain that the 95 ft.lbs spec applies to input torque,although it seems reasonable. The output torque to the hub becomes 95*1.87=177 ft.lbs.(quite massive).

It pays to derate mechanical devices especially expensive ones that rely on contact friction,if you expect them to last,and it would seem prudent to limit the torque input to below 70,corresponding to about 1.7 HP.(at 130 input rpm).Knowing the max torque from an engine it is not difficult to come up with a suitable max reduction,the Robin EH 035 has a max torque if 1.8 at 5k rpm,this implies a reduction of less than 70/1.8 or about 40 and the input to the hub would become 125 rpm.(at 5k engine rpm) This would leave some margin for pedal assist.It should be borne in mind that the average pedal torque is likely to be less than 50% of the peak torque and the device is peak limited.If larger, that is higher torque engine are used than the Robin the reduction ratio would have to be further reduced.You could go faster uphill but could not climb steeper ones,that may be academic unless you wanted to attack something like Pikes peak, 120 lbs pull (55kg).will lug over 400 lbs up a 25% grade!,if the wheel can take the punishment that is.

I would welcome inputs from people with NuVinci experience,esp. as regards durability of the device.In engineering: If you can't quantify it, you don't really know what you're talking about.

.