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 CB1100 not so unremarkable as I thought
#40
(08-07-2017, 11:12 AM)Randy B_imp Wrote:
(08-07-2017, 05:33 AM)Aka Tsubasa_imp Wrote:
(08-07-2017, 04:57 AM)Cormanus_imp Wrote:
(08-07-2017, 02:46 AM)Randy B_imp Wrote:
(08-07-2017, 12:06 AM)postoak_imp Wrote: Yeah, I'm still not getting it.

Here's something from Quora:

QuoteTongueutting some numbers to this, and plucking these numbers out of the air, if the engine is connected to a 50 tooth gear which is driving a 100 tooth gear which is connected to the gearbox output shaft then we have a 1:2 gear ratio, leading to the 100 tooth gear's shaft rotating at exactly half the speed of the engine but with twice as much torque. Since brake horsepower is simply (torque * revs), I'm simplifying slightly but this is close enough to illustrate the point, then the amount of power being delivered to the 100 toot gear's shaft is exactly the same as that which the engine is delivering to the 50 tooth gear-shaft: twice as much torque at half as many revs.
So, the way I see that is that the engine's horsepower is constant at any given rpm, but the torque can change at the wheels because of gearing.

So, the way I see that is that the engine's horsepower is constant at any given rpm, but the torque can change at the wheels because of gearing.
I think your issue is you are introducing a variable that has nothing to do with the power that the engine produces....gear ratios.
Do they effect the amount of power put to the ground? Absolutely. There are many varying gear ratios in the transmission and then the final drive, be it gears or a chain and sprockets.
I'm not talking about those for our purpose right now, just the engine because differing gear ratios produce differing amounts of torque multiplication depending on the amount of under or overdrive in them.
This can get really complicated, really fast and heads can spin out of control...lol.

Just take a look at the equation I posted and just think about the engine without and gears...no transmission or final drive. That is the simplest way to think about it.
It will make sense...at some point...lol.
Indeed. And, at the risk of being tediously repetitious, see [url=http://cb1100forum.com/forum/showthread.php?pid=180160#pid180160]here. Tongue
Aka Tsubasa, you say Torque=Force x Distance. Randy B says Torque=Force x Time. I suppose time and distance might be equal, but that isn't necessarily the case. So which of those statements is correct?

A certain late and lamented forum member would be helpful at this juncture.
Indeed. And, at the risk of being tediously repetitious, see [url=http://cb1100forum.com/forum/showthread.php?pid=180160#pid180160]here. Tongue
Aka Tsubasa, you say Torque=Force x Distance. Randy B says Torque=Force x Time. I suppose time and distance might be equal, but that isn't necessarily the case. So which of those statements is correct?

A certain late and lamented forum member would be helpful at this juncture.
Unless Physics has changed in the last decade, I am sure it is distance. Hence units such as lb-ft. Ft being distance.
Indeed. And, at the risk of being tediously repetitious, see [url=http://cb1100forum.com/forum/showthread.php?pid=180160#pid180160]here. Tongue
Aka Tsubasa, you say Torque=Force x Distance. Randy B says Torque=Force x Time. I suppose time and distance might be equal, but that isn't necessarily the case. So which of those statements is correct?

A certain late and lamented forum member would be helpful at this juncture.
Unless Physics has changed in the last decade, I am sure it is distance. Hence units such as lb-ft. Ft being distance. Actually ft-lb is the amount of torque applied 1 foot from the centerline of rotation. So picture a 50 lb weight hanging off the end of a 1' long bar. That will exert 50 ft-lbs of rotational force at the other end of the bar.
And RPM is a unit of time, not distance.
(08-07-2017, 04:57 AM)Cormanus_imp Wrote:
(08-07-2017, 02:46 AM)Randy B_imp Wrote:
(08-07-2017, 12:06 AM)postoak_imp Wrote: Yeah, I'm still not getting it.

Here's something from Quora:

QuoteTongueutting some numbers to this, and plucking these numbers out of the air, if the engine is connected to a 50 tooth gear which is driving a 100 tooth gear which is connected to the gearbox output shaft then we have a 1:2 gear ratio, leading to the 100 tooth gear's shaft rotating at exactly half the speed of the engine but with twice as much torque. Since brake horsepower is simply (torque * revs), I'm simplifying slightly but this is close enough to illustrate the point, then the amount of power being delivered to the 100 toot gear's shaft is exactly the same as that which the engine is delivering to the 50 tooth gear-shaft: twice as much torque at half as many revs.
So, the way I see that is that the engine's horsepower is constant at any given rpm, but the torque can change at the wheels because of gearing.

So, the way I see that is that the engine's horsepower is constant at any given rpm, but the torque can change at the wheels because of gearing.
I think your issue is you are introducing a variable that has nothing to do with the power that the engine produces....gear ratios.
Do they effect the amount of power put to the ground? Absolutely. There are many varying gear ratios in the transmission and then the final drive, be it gears or a chain and sprockets.
I'm not talking about those for our purpose right now, just the engine because differing gear ratios produce differing amounts of torque multiplication depending on the amount of under or overdrive in them.
This can get really complicated, really fast and heads can spin out of control...lol.

Just take a look at the equation I posted and just think about the engine without and gears...no transmission or final drive. That is the simplest way to think about it.
It will make sense...at some point...lol.
Indeed. And, at the risk of being tediously repetitious, see [url=http://cb1100forum.com/forum/showthread.php?pid=180160#pid180160]here. Tongue
Aka Tsubasa, you say Torque=Force x Distance. Randy B says Torque=Force x Time. I suppose time and distance might be equal, but that isn't necessarily the case. So which of those statements is correct?

A certain late and lamented forum member would be helpful at this juncture.

Actually I said horsepower is force (torque) x time (RPM)/5252. I never stated an equation for torque. Tongue

You need to get off your head man. Being upside down for so long is beginning to mess with you a little....lol. Excited ROFL

I keeed.....I keeeed..... Thumbs Up

Actually I said horsepower is force (torque) x time (RPM)/5252. I never stated an equation for torque. Tongue

You need to get off your head man. Being upside down for so long is beginning to mess with you a little....lol. Excited ROFL

I keeed.....I keeeed..... Thumbs Up
Apologies for misquoting you.

From where I usually stand, it's you who's upside down. Banana

Yet, when I make the pilgrimage to the other hemisphere, I find I'm still standing upright and I can see your perspective. Curious. Why is it so? Excited
Reply


Messages In This Thread
RE: CB1100 not so unremarkable as I thought - by Cormanus - 08-07-2017, 03:59 PM

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