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RE: CB1100 not so unremarkable as I thought - Randy B - 08-06-2017

(08-06-2017, 01:25 PM)Cormanus_imp Wrote: Please sing along with Cormanus ...

"I'm so dizzy my head is spinning
Like an oil thread, it never ends
And it's torque that's making it spin
It's making me dizzy."

Excited Celebrate Big Grin Hello ROFL ROFL ROFL ROFL Worship Thanks Thanks Thanks Thanks


RE: CB1100 not so unremarkable as I thought - Offroad - 08-06-2017

All I know is it goes fast enough for me ;-)


RE: CB1100 not so unremarkable as I thought - Blockhead_imp - 08-06-2017

Randy B, that is the best explanation of torque/horsepower that I've ever been able to process without my eyes glazing over... Thanks


RE: CB1100 not so unremarkable as I thought - paulw_imp - 08-06-2017

I remember seeing a funny racing saying some years ago that sort of said it all to me.

"Horsepower will determine how fast you hit the wall and torque will determine how much of the wall you take with you"


RE: CB1100 not so unremarkable as I thought - postoak_imp - 08-07-2017

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.


RE: CB1100 not so unremarkable as I thought - the Ferret - 08-07-2017

Sometimes trying to understand something complicated, takes the fun out of the thing you are trying to understand.

Ride more, worry less.


RE: CB1100 not so unremarkable as I thought - Randy B - 08-07-2017

(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.


RE: CB1100 not so unremarkable as I thought - Aka Tsubasa_imp - 08-07-2017

OK. I will take a stab at clearing this up.

Torque = Force x Distance (turning force)
Horsepower = How much torque is applied over time

Now, the super simple explanation:

Torque = Magic that makes stuff move
Horsepower = How quickly that magic can be applied

Bottom Line: A person wants as much torque as they can get at any given RPM becuase this results in the highest HP possible for that engine speed.


RE: CB1100 not so unremarkable as I thought - Cormanus - 08-07-2017

(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.


RE: CB1100 not so unremarkable as I thought - Aka Tsubasa_imp - 08-07-2017

(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.