CB1100 not so unremarkable as I thought

[Randy B]

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."

[offroadfx4_imp]

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

[Blockhead_imp]

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

[paulw_imp]

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"

[postoak_imp]

Yeah, I'm still not getting it.

Here's something from Quora:

Quoteutting 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.

[the Ferret]

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

Ride more, worry less.

[Randy B]

Yeah, I'm still not getting it.

Here's something from Quora:

Quoteutting 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.

[Aka Tsubasa_imp]

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.

[Cormanus]

Yeah, I'm still not getting it.

Here's something from Quora:

Quoteutting 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.
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.

[Aka Tsubasa_imp]

Yeah, I'm still not getting it.

Here's something from Quora:

Quoteutting 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.
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.
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.