loquin
Well-Known Member
Here's an online calculator. (As an alternative to this website, I've written an executable you can install on your computer that can be simpler to use, but with more options, available here.)
Essentially, it will calculate the power required, in watts, to maintain a given speed, with headwind/tailwind, on a given slope.
To convert from a slope to an angle, take the inverse tangent of the slope. 10% slope is 0.10, so the inverse tangent of 0.10 is 5.7 degrees. If your calculator is set for radians, either set it to degrees, or multiple the radians by 180, then divide by pi to get degrees. (Or, multiple radians by 57.3 degrees/radian.)
Here's an example. Suppose you + bike + motor + cargo weighs 300 pounds. You need to calculate how much power is needed to push the whole kaboodle up a 10 percent slope at 15 MPH. Per the power calculations at the link, above, you would need 931 watts, or, 1.25 horsepower. If your engine is to 'work' in this situation, you need to have it running in it's high output RPM range at the calculated speed, which forces the 'gear ratios' to be higher than you might otherwise prefer. This is also why a CVT can be a good alternative in hilly regions. Although you sacrifice some efficiency, and as a result, reduce the total system top end speed, you gain the ability to climb those hills sitting down...
Keep in mind that this is the power at the AXLE, not at the motor output shaft. If you have a belt drive, a chain drive, a gearbox, and.or a CVT (or any combination,) you would also need to account for the losses between the motor shaft and the axle.
As an example, suppose you wanted to calculate the power available at the axle from a R/S EHO 35 (1.6HP) engine, set up with a pocket bike CVT, Chain to a jackshaft, then chain to the axle.
You would multiply the 1.6HP times .92 times .98 * .98, to calculate that approximately 1.41 HP (1044 watts) would be available at the axle. Using our earlier assumptions for weight and speed, this motor would be able to push a three hundred pound load up a 7.1 degree (12.4%) slope at 15 MPH.
If you were looking at a more severe test, let's look at things with a 400 pound total load, and a 20 percent slope, using the same motor. The arc-tangent of 20% (.2) is 11.3 degrees. Adjusting the speed until the total power is approximately 1044 watts results in a speed estimate of 6.7 MPH. Since the engine max torque is achieved at about 5000 RPM, you could calculate the gearing which would return this speed at 5000 RPM, (including the CVT belt and gearbox reduction,) and then, when the apx. 2.2 belt reduction and increase to 7800 RPM is calculated, it could be shown that this motor could push a substantial load up a steep hill, and yet cruise at a maximum speed of about 23 MPH.
Since these speeds would only available when using a new belt, in this case, you would probably want to assume an 85% efficiency in the CVT. This would result in a usable power number at the axle of apx. 975 watts, and gearing calculated for maximum hill climbing speed of about 6.2 MPH at 5000 RPM, and a top speed (on the flat) of about 21 MPH.
Essentially, it will calculate the power required, in watts, to maintain a given speed, with headwind/tailwind, on a given slope.
To convert from a slope to an angle, take the inverse tangent of the slope. 10% slope is 0.10, so the inverse tangent of 0.10 is 5.7 degrees. If your calculator is set for radians, either set it to degrees, or multiple the radians by 180, then divide by pi to get degrees. (Or, multiple radians by 57.3 degrees/radian.)
Here's an example. Suppose you + bike + motor + cargo weighs 300 pounds. You need to calculate how much power is needed to push the whole kaboodle up a 10 percent slope at 15 MPH. Per the power calculations at the link, above, you would need 931 watts, or, 1.25 horsepower. If your engine is to 'work' in this situation, you need to have it running in it's high output RPM range at the calculated speed, which forces the 'gear ratios' to be higher than you might otherwise prefer. This is also why a CVT can be a good alternative in hilly regions. Although you sacrifice some efficiency, and as a result, reduce the total system top end speed, you gain the ability to climb those hills sitting down...
Keep in mind that this is the power at the AXLE, not at the motor output shaft. If you have a belt drive, a chain drive, a gearbox, and.or a CVT (or any combination,) you would also need to account for the losses between the motor shaft and the axle.
- A well-aligned & maintained chain is about 98% efficient. As the chain gets worn, it will stretch a bit, and you get slippage on the sprockets. (The chain roller will not be contacting the sprockets exactly in the center of the 'trough.' This leads to decreased efficiency, and increased wear. Also, misalignment of the sprockets causes a reduction in efficiency.
. - A toothed drive belt (GEBE) is about the same - about 98%. A toothed, synchronous belt MUST be aligned very well, else it tends to cause increased wear, decreased efficiency, and very short belt life.
. - A V-Belt can be fairly efficient as well - per PlantServices.com, "V-belt drive efficiency can run as high as 95 to 98% at the time of installation. During operation, however, V-belt efficiency deteriorates as much as five percent. The efficiency of a poorly maintained V-belt may fall an additional 10%."
. - Belt CVTs (Comet/Max-Torque) with a new belt start off life at about 95% efficiency with a new belt. As the belt wears over time, the efficiency slowly drops, to a low of about 85%.
. - In addition to the built-in belt drive, pocket bike CVTs also have a 3:1 gearbox on the output, which will also reduce efficiency a bit. So, you're probably looking at an output which starts off at about 92% or so, then slowly drops over time to about 82%. Put in a new belt, and the efficiency should shoot back up to the 92% range again.
. - A gearbox is variable - the number and type of gears will affect the efficiency. A rule of thumb I've seen used for efficiency, is 95%-98% per reduction for spur and helical gears. Using this, an 18.75:1 Staton gearbox, for instance, with it's three gears, and 2 gear-gear meshes, might be in the 90% to 96% efficiency range. If you split the difference, and assume 93%, you probably wouldn't be too far off. Worm gears can be much less efficient, losing anywhere from 2 to 80 percent (!!!) of the input power, depending upon their reduction ratio, gear angle, and gear material. Maximum efficiency occurs when the worm gear pressure angle is 45 degrees. Reference. Because of their potential enormous efficiency range, these gearboxes must be addressed on a case-by-case basis.
. - There IS no official Nuvinci published efficiency figures; at their forum, they never answer the questions about efficiency - they just evade them and refer you to tests done with it by professional riders and others, and discuss how the hub 'feels' and how it is a good 'total system' choice. The only third party data I saw indicated an efficiency of a little less than 90%.
. - Contrary to (some) opinion, friction drives can be fairly efficient. Essentially, their losses are another form of bike tire rolling friction; as the tire deflects around the roller, some force is needed to flex the tire. Although the amount of deflection depends upon the tire inflation pressure and down force applied when locking the roller into place, with a tire pressure of 40-50 psi, a reasonable amount would be not more than 5 times the standard rolling friction of a bike tire, (which is substantially less than 1 percent. So, an efficiency of 96% could be conservatively used in your calculations.
As an example, suppose you wanted to calculate the power available at the axle from a R/S EHO 35 (1.6HP) engine, set up with a pocket bike CVT, Chain to a jackshaft, then chain to the axle.
You would multiply the 1.6HP times .92 times .98 * .98, to calculate that approximately 1.41 HP (1044 watts) would be available at the axle. Using our earlier assumptions for weight and speed, this motor would be able to push a three hundred pound load up a 7.1 degree (12.4%) slope at 15 MPH.
If you were looking at a more severe test, let's look at things with a 400 pound total load, and a 20 percent slope, using the same motor. The arc-tangent of 20% (.2) is 11.3 degrees. Adjusting the speed until the total power is approximately 1044 watts results in a speed estimate of 6.7 MPH. Since the engine max torque is achieved at about 5000 RPM, you could calculate the gearing which would return this speed at 5000 RPM, (including the CVT belt and gearbox reduction,) and then, when the apx. 2.2 belt reduction and increase to 7800 RPM is calculated, it could be shown that this motor could push a substantial load up a steep hill, and yet cruise at a maximum speed of about 23 MPH.
Since these speeds would only available when using a new belt, in this case, you would probably want to assume an 85% efficiency in the CVT. This would result in a usable power number at the axle of apx. 975 watts, and gearing calculated for maximum hill climbing speed of about 6.2 MPH at 5000 RPM, and a top speed (on the flat) of about 21 MPH.
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