Finally installed 16T sprocket - Page 8 - Stromtrooper Forum : Suzuki V-Strom Motorcycle Forums
General V-Strom Discussion Talk about all things V-Strom not limited to just one of the above models

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post #71 of 82 Old 04-14-2019, 09:43 AM
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Originally Posted by realshelby View Post
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Originally Posted by Hogges View Post
Sorry V-Tom, this is incorrect. This is why talking about torque at the wheel gets people in trouble. At the same rpm you have the same power and power directly translates into acceleration at a given weight.

Here is a link which explains it pretty well I think, mixing scientific formulas with practical examples: https://danielmiessler.com/study/horsepower/

Are you sure V-Tom is incorrect? He is comparing a 15t to a 16t front sprocket. Don't confuse horsepower/torque calculations with this. Same bike/same conditions is what the debate is about. Comparing a 15 and 16 tooth sprocket.

Gearing ratios effect rear wheel torque. Mechanical advantage. You will have more rear wheel torque with a 15t compared to a 16t at that given rpm. That does NOT mean it does more work ( horsepower ) as that is calculated over time, not instant like torque. That advantage goes away when the 15t has to upshift sooner.
Let me grab a coffee and check this. I used to be pretty good in math and physics, back in school. Now work seems to mostly consist of answering emails and the phone... V-Tom dictated that we compare at the same rpm, which means at the same horsepower but at different speeds (because of the changed gearing). I’ll report back.
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post #72 of 82 Old 04-14-2019, 10:55 AM
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Ok here it goes, and a big “mea culpa” from me to V-Tom!
a=F/m and P=F*v. This means a=P/(m*v).
The acceleration is proportional to power, but also inverse proportional to the speed. It takes increasingly more power to accelerate at the same rate with increased speed.
In V-Tom’s scenario we are going 6% faster with the taller gearing, to achieve the same rpm. Because if this, acceleration will be necessary slower, as everything else (P, m) is unchanged.
The faster we go, the harder it is to accelerate from there.
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post #73 of 82 Old 04-14-2019, 10:59 AM
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Originally Posted by Hogges View Post
Sorry V-Tom, this is incorrect. This is why talking about torque at the wheel gets people in trouble. At the same rpm you have the same power and power directly translates into acceleration at a given weight.
...

If what you are saying is true then, at a given rpm we should accelerate the same rate whether we are in 1st gear or 6th gear. Doesn't seem to happen on my bike.

..Tom
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post #74 of 82 Old 04-14-2019, 12:06 PM
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Below is from your link. IF the torque at the rear wheels is a combination of engine output and magnification by the transmission then surely the final sprocket ratios play into that and taller gearing via a larger front sprocket should equal less rear wheel torque.

When bikes are measured on a Dyno they typically have to correct for gearing don't they? Meaning if you want to know what your engine is producing you have to figure in gear ratios and frictional losses to try to determine he crankshaft output. B

I think we are all kind of saying the same thing here.

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Gearing
So that’s where gearing comes in.

gearing magnifies torque. The torque at the wheels is the torque at the engine combined with the torque magnification given by the transmission through gearing. So the transmission only sees what’s coming off the engine, while the wheels see the resulting force combination of the engine plus the transmission.

That’s what horsepower represents. Horsepower is the combination of the benefits of the engine’s raw abilities combined with RPMs. And RPMs are what allow us to use gearing effectively, which gives us more torque at the wheels.



Quote:
Originally Posted by Hogges View Post
Sorry V-Tom, this is incorrect. This is why talking about torque at the wheel gets people in trouble. At the same rpm you have the same power and power directly translates into acceleration at a given weight.

Here is a link which explains it pretty well I think, mixing scientific formulas with practical examples: https://danielmiessler.com/study/horsepower/
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post #75 of 82 Old 04-14-2019, 12:54 PM
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Originally Posted by Hogges View Post
Sorry V-Tom, this is incorrect. This is why talking about torque at the wheel gets people in trouble. At the same rpm you have the same power and power directly translates into acceleration at a given weight.
...

If what you are saying is true then, at a given rpm we should accelerate the same rate whether we are in 1st gear or 6th gear. Doesn't seem to happen on my bike.

..Tom
Hi V-Tom. I already corrected my statement. Did not want to edit the original post to not confuse matters. My apologies. But of course in different gears we have different power, this is why we shift down to pass.
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post #76 of 82 Old 04-14-2019, 02:19 PM
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Question Gearing Question

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I went from the stock 15T sprocket with the rubber damper to a solid steel 16T and I can't tell any difference in noise or vibration from the chain. The engine is less buzzy at highway speeds. 70 to 80 mph.
So if I am reading the correctly I would need to increase the tooth count of the front sprocket or reduce the tooth count of the rear sprocket if my goal was to reduce engine RPM at highway speeds?

I don't think I have read of anyone doing that, can someone tell me how rear sprocket tooth counts relate to engine RPM. For instances if I dropped two teeth on the rear sprocket what kind of RPM reduction could I expect?

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post #77 of 82 Old 04-14-2019, 02:24 PM
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Originally Posted by Hogges View Post
Ok here it goes, and a big “mea culpa” from me to V-Tom!
a=F/m and P=F*v. This means a=P/(m*v).
The acceleration is proportional to power, but also inverse proportional to the speed. It takes increasingly more power to accelerate at the same rate with increased speed.
In V-Tom’s scenario we are going 6% faster with the taller gearing, to achieve the same rpm. Because if this, acceleration will be necessary slower, as everything else (P, m) is unchanged.
The faster we go, the harder it is to accelerate from there.
I wonder if all this would be true if done in a vacuum or any other condition that would remove wind resistance/drag?

2003 Zuki V-Strom 1000
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post #78 of 82 Old 04-14-2019, 04:18 PM
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Ok here it goes, and a big “mea culpa” from me to V-Tom!
a=F/m and P=F*v. This means a=P/(m*v).
The acceleration is proportional to power, but also inverse proportional to the speed. It takes increasingly more power to accelerate at the same rate with increased speed.
In V-Tom’s scenario we are going 6% faster with the taller gearing, to achieve the same rpm. Because if this, acceleration will be necessary slower, as everything else (P, m) is unchanged.
The faster we go, the harder it is to accelerate from there.
I wonder if all this would be true if done in a vacuum or any other condition that would remove wind resistance/drag?
The formulas I used are just ideal kinetic formulas ignoring resistances and drag. Even in vacuum it gets harder and harder to accelerate as speed increases. This is also why kinetic energy increases with the square of velocity and not linearly.

Adding air resistance will make this even worse as we use a large portion of the engine horsepower just to maintain our speed, and only what’s left for acceleration.
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post #79 of 82 Old 04-14-2019, 05:03 PM
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Dyno runs are usually done in the most direct ratio ( ie 1 to 1 ) available. There is less frictional loss in a direct drive. Taller gears also tend to have less tire slippage on the drum too.

But you can run any gear ratio in the rear end, and make the dyno run in any gear.

The results will be the same, barring any losses coming from gear mesh and efficiency ( parasitic drag ). And there will be a few hp less in different gears and with different gearing ratios. Again, due to frictional losses.

The dyno uses drum rpm to calculate torque. It senses how quick the drum picks up speed. With taller gears, the dyno run starts at higher rpm vs lower gears assuming both runs start at 3000 engine rpm. The dyno drum is turning faster with the taller gearing when both runs are stopped at the same engine rpm. That is how the dyno allows for different gearing and still gives more or less the same reading for the same engine. The dyno computer can calculate power from there.
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post #80 of 82 Old 04-14-2019, 06:57 PM
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So if I am reading the correctly I would need to increase the tooth count of the front sprocket or reduce the tooth count of the rear sprocket if my goal was to reduce engine RPM at highway speeds?
Yes, assuming that the gear number is unchanged, those tooth count changes will reduce engine RPM (at any given speed).

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I don't think I have read of anyone doing that, can someone tell me how rear sprocket tooth counts relate to engine RPM. For instances if I dropped two teeth on the rear sprocket what kind of RPM reduction could I expect?
From measurement and inspection of the geometry and ratios from the service manual:
(Using left-associative multiply and divide)
Vmph = (Erpm rotations/minute) / Npr / Ntr[gear] * (Nf/Nr) * (Rrear inches) * 2*Pi * (60 minutes/hour) / (5280*12 inches/mile)
where
Vmph := bike speed in miles per hour
Erpm := engine speed in rotations per minute
Npr := primary reduction ratio (71/34 for V-Stroms)
Ntr[1..6] = { 32/13, 32/18, 29/21, 27/24, 25/26, 23/27 } // a tuple of 6 transmission ratio numbers, indexed by ...
gear := an integer between 1 and 6 inclusive representing the selected gear
Nf := tooth count of front sprocket (15 stock)
Nr := tooth count of rear sprocket (47 stock)
Rrear := effective radius of rear tire (distance from rear contact patch to loaded rear axle center) in inches

Combining the constants, including Npr, (and ditching units for checking):
Vmph = Erpm / Ntr[gear] * (Nf/Nr) * Rrear / 351

So, for example, on my bike (as now sprocketed) in 6th gear at 5k RPM, with Rrear = 12.2:
Vmph = 5000 / (23/27) * (16/47) * 12.2 / 351 = 69.5 miles/hour (actual)

Or in 4th gear at 2500 rpm, when it had the stock front sprocket:
Vmph = 2500 / (27/24) * (15/47) * 12.2 / 351 = 24.7 miles/hour (actual)
That lower speed I know well because my GPS would indicate 25 mph at that rpm and gear.
(I often went through a 25 MPH, speed-trapped neighborhood, so learned to keep RPM at 100 desired speed in MPH.)

Another example I remember well from the break-in period:
Vmph = 5000 / (23/27) * (15/47) * 12.2 / 351 = 65.1 miles/hour (actual) or 62.7/0.93 = 70.0 (indicated)

The 0.93 divisor to get from actual to indicated speed reflects the speedometer reading 7 percent low.
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