(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."
Tommy Roe has a CB1100. Awesome.
(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:
Quote
utting 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.
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. 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:
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.
Actually I said horsepower is force (torque) x time (RPM)/5252. I never stated an equation for torque.
You need to get off your head man. Being upside down for so long is beginning to mess with you a little....lol.
I keeed.....I keeeed.....
I'm surprised that this thread about horsepower and torque has lasted as long as it has considering how little the CB has of both.
(08-07-2017, 12:19 PM)Frulk_imp Wrote: [ -> ]I'm surprised that this thread about horsepower and torque has lasted as long as it has considering how little the CB has of both.
"Like"
Lol what is horsepower, how much is needed, what is torque, what 's it's roll in powering a motorcycle is always a hot topic, on every site I have ever been on.
The CB only has a little horsepower and torque compared to motorcycles which have more. Compared to say a V7 Guzzi the CB1100 is a horsepower and torque monster lol.
The CB has plenty of torque. Not compared to some big V twins or a diesel tractor but if you compare her numbers to a host of other bikes she comes out pretty good.
Work=Force * distance. Maybe that is where the earlier poster thought it came in.
Force=mass*acceleration.
Then you can say Work=mass*acceleration*distance - think drag racer. Also works better for rockets, like the Shuttle.
Naturally the numbers will change depending upon the oil you use.
Well, I hate to beat a dead duck, but torque is what allowed me to beat my buddy stoplight to 10 over the speed limit on my SR500. He had a Honda 400f.
I really don't care about the numbers. I never ride much over 5k rpm so I will probably never get to the top of this bikes hp curve anyway. Comparisons are odious. If you aren't happy with the way the machine moves your butt down the road, buy something else. There are a lot of choices.
Ben
Those concerned about all out engine performance (or perhaps performance in general, lol) should just go ahead and check their CB1100s at the door. There are plenty of bikes that can outperform the CB in real life and on paper for a lot less money. If raw performance is someone's main point of measurement for satisfaction, then I can't imagine how they would end up with a CB1100.
Obviously it's in our nature to geek out on this kind of stuff. For motorheads horsepower and torque figures that are discussed and dissected in any number of ways just like baseball fans fixate on batting averages and ERAs. In either case, be it motorcycles or baseball players, rarely are my favorites at the top of the stats heap. There are many factors other than these primary performance statistics that determine the true character of a motorcycle or the makeup of a ballplayer.
If we're only focusing on engine characteristics (as opposed to handling characteristics, etc.) and tossing aside bragging rights (be it for raw performance, or fuel efficiency, or that have you), I'd say that when it comes to motorcycle engines, this is what matters the most in reality: is it tuned in such a way that makes it truly enjoyable given the way you prefer to ride? Pretty simple.
(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:
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.
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. 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:
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.
Actually I said horsepower is force (torque) x time (RPM)/5252. I never stated an equation for torque.
You need to get off your head man. Being upside down for so long is beginning to mess with you a little....lol.
I keeed.....I keeeed.....
Actually I said horsepower is force (torque) x time (RPM)/5252. I never stated an equation for torque.
You need to get off your head man. Being upside down for so long is beginning to mess with you a little....lol.
I keeed.....I keeeed.....
Apologies for misquoting you.
From where I usually stand, it's you who's upside down.
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?