Why bike part manufactures doesn't make robust ebike specific drivetrain parts?

mt247

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Aug 12, 2020
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Middrives are very popular now days and it made me wonder why Shimano/Sram etc. doesn't make proper really robust drivetrain parts for ebikes? Wider rear sprockets, chainrings and chains.
 

soundwave

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May 23, 2015
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if you keep it clean and lubed and indexed it will last that is my 2nd cassette and sold the old 11spd one on ebay with 1000s of miles on it and i use a dongle ;)



DSC_0096.JPG
 

vfr400

Esteemed Pedelecer
Jun 12, 2011
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Weight is not issue with Ebikes though. You don't notice if your cassette + chain are in total 1kg heavier.
I would. I already pay extra for lighter cranks, stem and gears. after that, there has to be a clear benefit for the extra weight, like pucture proof tyres and a sprung seat.

Apart from that, there hasn't been enough demand foe special ebike parts, so most cycle parts are carried over from non-electric bikes. That will eventually change in the future.

Also, you can always get a hub-motor, that gives you a cheaper, lighter, more reliable and more user-friendly drive system, then you can use the cheapest or lightest drive train components you want.
 

soundwave

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May 23, 2015
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;)
 

Ocsid

Esteemed Pedelecer
Aug 2, 2017
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Middrives are very popular now days and it made me wonder why Shimano/Sram etc. doesn't make proper really robust drivetrain parts for ebikes? Wider rear sprockets, chainrings and chains.
Probably because a European powered & speed legal e-bike's components should not be subjected to much higher load carrying loads than a non e-bike in a serious riders hands?
And probably the durability achieved by Mr & Ms Average, is tolerated in that market?

Wear issues, as with normal bikes comes more into focus where the envelope is pushed in the inflicted loading, abuse and/or the bike's amount of usage.
 
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Edward Elizabeth

Pedelecer
Aug 10, 2020
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Buckinghamshire
Weight, expense, unnecessarily not-cross compqtible with regular drivertrain components, etc.

Shimano are a very successful outfit, and I'm sure if there were a viable business case for it they'd be on it in an instant.
 
  • Agree
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mt247

Pedelecer
Aug 12, 2020
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Probably because a European powered & speed legal e-bike's components should not be subjected to much higher load carrying loads than a non e-bike in a serious riders hands?
And probably the durability achieved by Mr & Ms Average, is tolerated in that market?

Wear issues, as with normal bikes comes more into focus where the envelope is pushed in the inflicted loading, abuse and/or the bike's amount of usage.
Most of the 250W mid drives gives you 80-90nm of torque and that is a lot and normal bike drivetrains aren't designed for that kind of power.
 

mike killay

Esteemed Pedelecer
Feb 17, 2011
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Most of the 250W mid drives gives you 80-90nm of torque and that is a lot and normal bike drivetrains aren't designed for that kind of power.
It is not a matter of what they are designed for, but what is their safe working load.
 

Sturmey

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Jan 26, 2018
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Dont forget that a good athletic cyclist can output well over 250w and up to 400 mechanical watts. It would take over 500 electrical watts to equal this.
 

Ocsid

Esteemed Pedelecer
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It is worth noting that even a 56 kg rider using 165 mm cranks and standing on the pedals develops the quoted 90 NM of crank torque.

That's just with a softly applied load, it's going to be way more if the rider “pumps” their weight when pedalling hard, not to mention if they bring in direct muscle “power” by say using cleats.

And of course, the bike designers IMO would have in mind that some riders might be heavier than 56 kgs
 

Nealh

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For 400w human power one needs to be a pro cyclist, 400w isn't far off the average of a Tdf rider with bursts up to 1300 -1400w for the sprints.
 

mt247

Pedelecer
Aug 12, 2020
111
17
Dont forget that a good athletic cyclist can output well over 250w and up to 400 mechanical watts. It would take over 500 electrical watts to equal this.
Pro cyclist average between 200-300 watts. Your average cyclists are far below that.
 

mt247

Pedelecer
Aug 12, 2020
111
17
It is worth noting that even a 56 kg rider using 165 mm cranks and standing on the pedals develops the quoted 90 NM of crank torque.

That's just with a softly applied load, it's going to be way more if the rider “pumps” their weight when pedalling hard, not to mention if they bring in direct muscle “power” by say using cleats.

And of course, the bike designers IMO would have in mind that some riders might be heavier than 56 kgs
Torque applied to the crank is not the same as torque in the drivetrain.


If a device is rotating at a steady rate or is stationary, the torques applied to it must add up to zero -- any torque applied at one point must be taken off at another. So, with a bicycle crankset, the torque applied at the pedals is equal and opposite that taken off by the chain, neglecting the small amount lost in friction.
Let's look at a bicycle drivetrain starting with the cyclist's feet. Torque is conveyed from a pedal through the crank -- and from the left crank, then also through the bottom-bracket spindle -- to the chainwheel. Generally, the cyclist's rising leg applies a light torque opposite that of the descending leg. The chain, at the chainwheel, produces a torque equal and opposite the sum of the torques applied at the pedals.
Let's put some numbers to this We assume that the left leg is descending in mid-stroke, applying a force of 100 pounds directly downward on the left pedal, while the rising right leg is applying a reverse force of 10 pounds to the right pedal. Crank length is usually given in millimeters, but we're using English measurement here, as it's more familiar to most English-speaking readers. We'll assume 0.56 foot (170 mm) cranks.
The torque on the bottom-bracket spindle is 56 pound-feet: the 100-pound force at the pedal, times the 0.56 foot length of the crank. The torque at the chainwheel is slightly less, 50.4 pound feet, after we subtract the -5.6 pound-foot torque from the right pedal.
Now, let's assume a 50-tooth chainwheel. This has a radius of about 4 inches, or 0.33 feet. We can now calculate the chain tension:
50.4 pound-feet/0.33 feet = 153 pounds.
We'll assume a 20-tooth sprocket, with a radius of 1.6 inches -- 0.13 foot at the rear wheel. The 153-pound chain tension produces a 20.2 pound-foot torque at this sprocket:
153 pounds * 0.13 foot = 20.2 pound-feet.
(We could also simply note the ratio of tooth counts of the chainwheel and sprocket, 50/20, which would give us the same torque ratio.)
We'll also assume a 28-622 rear tire, with a nominal diameter of 26.76 inches and radius, 13.38 inches, or 1.11 feet.
We can now calculate the drive force where the rubber meets the road:
20.2 pound-feet / 1.11 feet = 18.2 pounds.
The ratio of the force at the pedals to that at the road is 90 : 18.2, or 4.95.
In these calculations, we neglect forces which would not contribute to torque: pedal force not in the direction of rotation, and the weight on the rear wheel. We also neglect friction, which reduces the drive force by a few percent. Numbers are rounded -- close, but not exact.
The ratio of the force at the pedals to drive force at the road is the gain ratio, which can be calculated more simply as the ratio of road speed to pedal speed, like this:
.(1.11 feet /.56 feet) * 50 teeth/20 teeth = 4.95
This calculation is simple and elegant, but our torque calculations give us some additional useful results:
  • the torque on the bottom-bracket axle
  • the effect of the reverse force from the cyclist's rising leg,
  • and the tension on the chain.
 

soundwave

Esteemed Pedelecer
May 23, 2015
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:p
15.5mph o_O