I need your help- HP vs torque discussion
02-19-2009, 02:11 PM
#61
Rennlist Member
Thread Starter
You are not transfering full torque either, due to wheel spin. the net force of acceleration at that speed would be the HP put to the pavement.
Even at a clutch released speed of say, your speed mentioned of "35mph". you are now at max torque, and a fraction of max HP. (creating an acceleration of A =F/M. But since A = p/(m x v) if you can increase power, you could get a faster acceleration, right? so, since tires are desiged to grip at slip levels, you could find a way to get wheel spin at that speed, to take advatage of greater HP levels.
Ever notice when some dragsters "hook up" at slow speeds, they bog and loose the race? They need some wheel spin to create the greatest acceleration rates. There is an optimal amout of wheel spin needed to transfer the most amout of HP possible.
mk
Even at a clutch released speed of say, your speed mentioned of "35mph". you are now at max torque, and a fraction of max HP. (creating an acceleration of A =F/M. But since A = p/(m x v) if you can increase power, you could get a faster acceleration, right? so, since tires are desiged to grip at slip levels, you could find a way to get wheel spin at that speed, to take advatage of greater HP levels.
Ever notice when some dragsters "hook up" at slow speeds, they bog and loose the race? They need some wheel spin to create the greatest acceleration rates. There is an optimal amout of wheel spin needed to transfer the most amout of HP possible.
mk
you are correct in that you are not transferring the full Power to the wheels but you are transfering the full Torque. IF the torque output of the engine did not match the torque put into the clutch your engine would acceterate (increase in RPM) or decellerate (decrease in RPM).
02-19-2009, 02:56 PM
#62
Rennlist Member
First let me mention that I never meant to argue the point that max HP is the ideal running point on a road course. I was merely trying to point out one scenario where torque output was the limiting factor not HP out, but and I agree this is a limited case is not what you would see on a road course.
Now to try to answer a few questions: I will mention a couple points then try to use examples to illustrate my points. The clutch slip and tire slip are all losses they will not gain you acceleration. The clutch does not effectively give you a different gear ratio. And engine torque out = trans mission torque in.
Tire slip: lets look at the burn out/brake torque, you are setting in first gear heal-toe on the brake and gas and engine racing at 6200 RPM and ou release the clutch. Now your billowing smoke off the two rear tires and your acceleration rate is all of a big aft 0 ft/s^2 . So do you have torque at the rear wheels? Yep, what ever the engine is generating multiplied by the gear ratios. How about HP? Yep. Usable HP? No. no work is being done propelling you car down the road. Why? it is being dissipated at the tires in the form of heat and noise.
Clutch slip: Now your setting stopped on a hill at a red light. Light turns green and you are about to go. You move your foot from the brake to the gas and start to let the clutch up when you see the idiot coming from the left getting ready to run the light. So waiting for them to get through you are momentarily riding the clutch keeping you from rolling back into the car behind. At this moment you must have torque at the rear wheels or gravity would be pulling you back wards, but again you are not accelerating and you engine must be generating some HP as it’s running. Again this power is dissipated at the slipping clutch in the form of heat. So torque is transmitted but power is not.
Clutch gear ratio? The clutch does not effectively give you a different gear ratio. If this where the case you would transmit power through the clutch and torque and RPM would transmit through using a multiplies of T*Gear ratio and RPM*(1/GR). The easiest one to see here is that there is no direct relation ship between Engine RPM and Trans RPM. Also from the previous example you can see that torque is transmitted through, and that power (some if not all) is disipated.
Now to try to answer a few questions: I will mention a couple points then try to use examples to illustrate my points. The clutch slip and tire slip are all losses they will not gain you acceleration. The clutch does not effectively give you a different gear ratio. And engine torque out = trans mission torque in.
Tire slip: lets look at the burn out/brake torque, you are setting in first gear heal-toe on the brake and gas and engine racing at 6200 RPM and ou release the clutch. Now your billowing smoke off the two rear tires and your acceleration rate is all of a big aft 0 ft/s^2 . So do you have torque at the rear wheels? Yep, what ever the engine is generating multiplied by the gear ratios. How about HP? Yep. Usable HP? No. no work is being done propelling you car down the road. Why? it is being dissipated at the tires in the form of heat and noise.
Clutch slip: Now your setting stopped on a hill at a red light. Light turns green and you are about to go. You move your foot from the brake to the gas and start to let the clutch up when you see the idiot coming from the left getting ready to run the light. So waiting for them to get through you are momentarily riding the clutch keeping you from rolling back into the car behind. At this moment you must have torque at the rear wheels or gravity would be pulling you back wards, but again you are not accelerating and you engine must be generating some HP as it’s running. Again this power is dissipated at the slipping clutch in the form of heat. So torque is transmitted but power is not.
Clutch gear ratio? The clutch does not effectively give you a different gear ratio. If this where the case you would transmit power through the clutch and torque and RPM would transmit through using a multiplies of T*Gear ratio and RPM*(1/GR). The easiest one to see here is that there is no direct relation ship between Engine RPM and Trans RPM. Also from the previous example you can see that torque is transmitted through, and that power (some if not all) is disipated.
02-19-2009, 05:49 PM
#63
Rennlist Member
Thread Starter
Good so we agree on the HP portion.
I think you maybe right about the clutch losses, but I think auto transmissions may deal with the "slip" differently. Someone with thoughts there can chime in, but its a different question. we are talking about slipping clutch only. Ok.
I agree, that the torque to the wheels when the clutch is released will follow the torque curve. (which is limted to the acceleration provided by the HP at that whatever speed that is. I know semantics at this point.)
Keep in mind, if the tires are spinning, there is a different level of force applied to the road. spinning wildly, just over the grip speed, and grip speed are all greatly different net forces applied to the car at the pavement. ft-lbs of torque can equal Lbs of thrust, if the tires are 24" in diameter. Use this if you need to so we can use wheel torque and thrust force interchangibly.
Let me inject a few comments below with the >>>>>>
Mk
I think you maybe right about the clutch losses, but I think auto transmissions may deal with the "slip" differently. Someone with thoughts there can chime in, but its a different question. we are talking about slipping clutch only. Ok.
I agree, that the torque to the wheels when the clutch is released will follow the torque curve. (which is limted to the acceleration provided by the HP at that whatever speed that is. I know semantics at this point.)
Keep in mind, if the tires are spinning, there is a different level of force applied to the road. spinning wildly, just over the grip speed, and grip speed are all greatly different net forces applied to the car at the pavement. ft-lbs of torque can equal Lbs of thrust, if the tires are 24" in diameter. Use this if you need to so we can use wheel torque and thrust force interchangibly.
Let me inject a few comments below with the >>>>>>
Mk
First let me mention that I never meant to argue the point that max HP is the ideal running point on a road course. I was merely trying to point out one scenario where torque output was the limiting factor not HP out, but and I agree this is a limited case is not what you would see on a road course.
Now to try to answer a few questions: I will mention a couple points then try to use examples to illustrate my points. The clutch slip and tire slip are all losses they will not gain you acceleration. The clutch does not effectively give you a different gear ratio. And engine torque out = trans mission torque in.
>>>>>>I agree with you here. It might be a different set of circumstances in autobox, but I agree it is as you say on this point.
Tire slip: lets look at the burn out/brake torque, you are setting in first gear heal-toe on the brake and gas and engine racing at 6200 RPM and ou release the clutch. Now your billowing smoke off the two rear tires and your acceleration rate is all of a big aft 0 ft/s^2 . So do you have torque at the rear wheels? Yep, what ever the engine is generating multiplied by the gear ratios. How about HP? Yep. Usable HP? No. no work is being done propelling you car down the road. Why? it is being dissipated at the tires in the form of heat and noise.
>>>>>>>>right, you have no net force (forces are equal with brake and clutch released and engine reving to 6k). HP applied is "0", but there is HP being wasted with the 6200rpm, and friction of the tires on the road making smoke. HP wasted is probably small and smaller as the tire rubber melts and the friction becomes less
However, at max torque (certain HP) and some rpm higher, there is more HP and if you have maximized slip of the tires, you will put more at the wheel torque (net force) and have more acceleration to the pavement. It would be because you are using more of the HP for acceleration than if you were at max torque. I dont know the exact percentages, but at some point pasted direct coupling with racing tires, you generate more G loading, both in cornering and in acceleration.
Clutch slip: Now your setting stopped on a hill at a red light. Light turns green and you are about to go. You move your foot from the brake to the gas and start to let the clutch up when you see the idiot coming from the left getting ready to run the light. So waiting for them to get through you are momentarily riding the clutch keeping you from rolling back into the car behind. At this moment you must have torque at the rear wheels or gravity would be pulling you back wards, but again you are not accelerating and you engine must be generating some HP as it’s running. Again this power is dissipated at the slipping clutch in the form of heat. So torque is transmitted but power is not.
<<<<<<<Great example, but dont confuse net force or power, with inefficiencies. Like sitting in a chair. two forces make a net force of 0, no work done, not power used. a chair is 100% efficient in creating this force. the car is less than 1% efficient holding the car up on the hill. Almost all the power is going up in heat.
Clutch gear ratio? The clutch does not effectively give you a different gear ratio. If this where the case you would transmit power through the clutch and torque and RPM would transmit through using a multiplies of T*Gear ratio and RPM*(1/GR). The easiest one to see here is that there is no direct relation ship between Engine RPM and Trans RPM. Also from the previous example you can see that torque is transmitted through, and that power (some if not all) is disipated.
>>>>>>>>> I agree, but I think we have to look at slip angles of the tires to see why max torque would not be the maximum acceleration in any gear in reality, as you could climb to a higher HP range to gain even more rear wheel torque, even though your engine torque would be slightly less.
It has to be this way, because there is a law that dictates acceleration.
Acceleration =power/(mass x velocity) it means at any speed, acceleration is directly proportional to power and inversely proportional to speed.
Now to try to answer a few questions: I will mention a couple points then try to use examples to illustrate my points. The clutch slip and tire slip are all losses they will not gain you acceleration. The clutch does not effectively give you a different gear ratio. And engine torque out = trans mission torque in.
>>>>>>I agree with you here. It might be a different set of circumstances in autobox, but I agree it is as you say on this point.
Tire slip: lets look at the burn out/brake torque, you are setting in first gear heal-toe on the brake and gas and engine racing at 6200 RPM and ou release the clutch. Now your billowing smoke off the two rear tires and your acceleration rate is all of a big aft 0 ft/s^2 . So do you have torque at the rear wheels? Yep, what ever the engine is generating multiplied by the gear ratios. How about HP? Yep. Usable HP? No. no work is being done propelling you car down the road. Why? it is being dissipated at the tires in the form of heat and noise.
>>>>>>>>right, you have no net force (forces are equal with brake and clutch released and engine reving to 6k). HP applied is "0", but there is HP being wasted with the 6200rpm, and friction of the tires on the road making smoke. HP wasted is probably small and smaller as the tire rubber melts and the friction becomes less
However, at max torque (certain HP) and some rpm higher, there is more HP and if you have maximized slip of the tires, you will put more at the wheel torque (net force) and have more acceleration to the pavement. It would be because you are using more of the HP for acceleration than if you were at max torque. I dont know the exact percentages, but at some point pasted direct coupling with racing tires, you generate more G loading, both in cornering and in acceleration.
Clutch slip: Now your setting stopped on a hill at a red light. Light turns green and you are about to go. You move your foot from the brake to the gas and start to let the clutch up when you see the idiot coming from the left getting ready to run the light. So waiting for them to get through you are momentarily riding the clutch keeping you from rolling back into the car behind. At this moment you must have torque at the rear wheels or gravity would be pulling you back wards, but again you are not accelerating and you engine must be generating some HP as it’s running. Again this power is dissipated at the slipping clutch in the form of heat. So torque is transmitted but power is not.
<<<<<<<Great example, but dont confuse net force or power, with inefficiencies. Like sitting in a chair. two forces make a net force of 0, no work done, not power used. a chair is 100% efficient in creating this force. the car is less than 1% efficient holding the car up on the hill. Almost all the power is going up in heat.
Clutch gear ratio? The clutch does not effectively give you a different gear ratio. If this where the case you would transmit power through the clutch and torque and RPM would transmit through using a multiplies of T*Gear ratio and RPM*(1/GR). The easiest one to see here is that there is no direct relation ship between Engine RPM and Trans RPM. Also from the previous example you can see that torque is transmitted through, and that power (some if not all) is disipated.
>>>>>>>>> I agree, but I think we have to look at slip angles of the tires to see why max torque would not be the maximum acceleration in any gear in reality, as you could climb to a higher HP range to gain even more rear wheel torque, even though your engine torque would be slightly less.
It has to be this way, because there is a law that dictates acceleration.
Acceleration =power/(mass x velocity) it means at any speed, acceleration is directly proportional to power and inversely proportional to speed.
02-19-2009, 06:37 PM
#64
Rennlist Member
I see you are really stuck on this slip angle issue. Slip angle does not increase thrust, it is a byproduct of it and in acceleration and braking I believe it is termed slip % not slip angle. Try thinking of it this way we'll pick some random numbers here 100lb*ft torque 100HP total gear ratio =1 and a 24" tire. What is the max thrust you can have with this set-up? 100 lb thrust, regardless what slip % you have the answer for max thrust is 100 lb. what the slip % does is give you a velocity offset directly proportional to the slip % itself. Below is a graph showing what a typical driving force vs. skip % may look like.
So from this you can see as long as you have not exceeded the 100% grip point the more thrust you generate the greater the slip angle will be. And as I believe we both agree the torque at the rear wheels it what generates that driving force. So max torque = max acceleration.
Slip % and Slip angle are not desirable they are merely a byproduct of tire design. if slip could be eliminated efficiency would increase. Imagine if you could use a rack and pinion with a 1/4 mile rack for drag racing... no slip all acceleration. Slip is parasitic
So from this you can see as long as you have not exceeded the 100% grip point the more thrust you generate the greater the slip angle will be. And as I believe we both agree the torque at the rear wheels it what generates that driving force. So max torque = max acceleration.
Slip % and Slip angle are not desirable they are merely a byproduct of tire design. if slip could be eliminated efficiency would increase. Imagine if you could use a rack and pinion with a 1/4 mile rack for drag racing... no slip all acceleration. Slip is parasitic
02-19-2009, 06:48 PM
#65
Race Car
I am 100% with you, great explanations. Maximum acceleration is attained in 1st gear if you keep the engine as close to the torque peak for as long as possible. In all other gears, maximum HP is the measure of acceleration. Now, what about "torque multiplication" on an automatic gearbox... ?
Dan
'91 928GT S/C
475hp/460lb.ft
Dan
'91 928GT S/C
475hp/460lb.ft
02-19-2009, 07:24 PM
#66
Rennlist Member
Thread Starter
Now, remember, the 100% of the braking force, is achieved at some X% slip, just as the 100% of the acceleration force is also at x%. Whatever the max potential power available is, at any speed, it will be achieved at some X% slip. agreed?
Now, the graphs are not saying that if the max potential of engine torque is 100ft.bs, it will be reached at this X% slip. Its saying that at X% slip, the max acceleration force will be available here.. Notice how the % of driving force is constant at 100% 200% slip? the torque at those speeds will be going down tremendously, but yet the acceleration % force is constant. Curious. I might not be interpreting the graph right, but is this only for a standing start? as I dont understand why the graph would be 0 % drving force at 0 slip.
Yes, I was talking about slip, when I said slip angle (a term used in corning acceleration)
It seems logical that a car at max torque accelerating at 100ft-lbs at what ever engine speed that is found, that it could accelerate harder and have more force at the wheels, even though the engine speed might be higher and the engine torque might be lower or even the same. (ie rising hp). I think this graph is kind of confusing, but maybe you can help clear it up, after all , its your graph.
The way I am interpreting the graph, (aside from the 0 slip% = 0% driving force) is that whatever the potential power avalable, you will only get a percentage of the acceleration if you dont get a little slip. This little slip, provides the ability of the vehicle to utilize more of the available HP, up to 100% of max power. Remember, Force = power x velocity.
Look at the braking force. it show 75% force, even though the brakes are locked. when they are locked, there is no braking force (on the brake system analogous to the engine), yet there is 75% braking force on the ground and tires available.
It shows me the % available force is more of total potential available at any speed.
This is a tough one. Maybe only for me. Looking forward to your response.
mk
Now, the graphs are not saying that if the max potential of engine torque is 100ft.bs, it will be reached at this X% slip. Its saying that at X% slip, the max acceleration force will be available here.. Notice how the % of driving force is constant at 100% 200% slip? the torque at those speeds will be going down tremendously, but yet the acceleration % force is constant. Curious. I might not be interpreting the graph right, but is this only for a standing start? as I dont understand why the graph would be 0 % drving force at 0 slip.
Yes, I was talking about slip, when I said slip angle (a term used in corning acceleration)
It seems logical that a car at max torque accelerating at 100ft-lbs at what ever engine speed that is found, that it could accelerate harder and have more force at the wheels, even though the engine speed might be higher and the engine torque might be lower or even the same. (ie rising hp). I think this graph is kind of confusing, but maybe you can help clear it up, after all , its your graph.
The way I am interpreting the graph, (aside from the 0 slip% = 0% driving force) is that whatever the potential power avalable, you will only get a percentage of the acceleration if you dont get a little slip. This little slip, provides the ability of the vehicle to utilize more of the available HP, up to 100% of max power. Remember, Force = power x velocity.
Look at the braking force. it show 75% force, even though the brakes are locked. when they are locked, there is no braking force (on the brake system analogous to the engine), yet there is 75% braking force on the ground and tires available.
It shows me the % available force is more of total potential available at any speed.
This is a tough one. Maybe only for me.
Looking forward to your response.mk
I see you are really stuck on this slip angle issue. Slip angle does not increase thrust, it is a byproduct of it and in acceleration and braking I believe it is termed slip % not slip angle. Try thinking of it this way we'll pick some random numbers here 100lb*ft torque 100HP total gear ratio =1 and a 24" tire. What is the max thrust you can have with this set-up? 100 lb thrust, regardless what slip % you have the answer for max thrust is 100 lb. what the slip % does is give you a velocity offset directly proportional to the slip % itself. Below is a graph showing what a typical driving force vs. skip % may look like.
So from this you can see as long as you have not exceeded the 100% grip point the more thrust you generate the greater the slip angle will be. And as I believe we both agree the torque at the rear wheels it what generates that driving force. So max torque = max acceleration.
Slip % and Slip angle are not desirable they are merely a byproduct of tire design. if slip could be eliminated efficiency would increase. Imagine if you could use a rack and pinion with a 1/4 mile rack for drag racing... no slip all acceleration. Slip is parasitic
So from this you can see as long as you have not exceeded the 100% grip point the more thrust you generate the greater the slip angle will be. And as I believe we both agree the torque at the rear wheels it what generates that driving force. So max torque = max acceleration.
Slip % and Slip angle are not desirable they are merely a byproduct of tire design. if slip could be eliminated efficiency would increase. Imagine if you could use a rack and pinion with a 1/4 mile rack for drag racing... no slip all acceleration. Slip is parasitic
02-19-2009, 07:49 PM
#67
Rennlist Member
Thread Starter
max HP at any speed will determine acceleration, but its a factor at launch how much HP the road can absorb. (ie wheel spin and optimal slip %)
I think a better analogy might be paddles in the water. sure, rack and pinion with a paddle hitting the rack and pushing, would be ideal. But, the faster the paddle moves, the more water it displaces and the greater the force. The friction of tires on the ground as well as the slip % acts in a viscous kind of way. No slip, and you dont maximize your acceleration potential at any speed.
If this is true, then theoretically, you migh be able to generate more force at a slightly higher rpm than at max torque rpm. In the end, this is due to having utilized more available HP. Its one of those things that I know the reason for, but trying to understand how it fits the problem. We all know, at any speed Power determines acceleration and the rear wheel torque. Anyone else want to give it a shot? Its the same reason why airplanes dont accelerate fastest at max torque. max HP always, absorbed by the propeller, provides the greatest thrust. (not max torque of the airplane's engine). if we are talking a motorboat, same thing. However, its much different with tires and the road, due to the slip of the rubber tire. utilizing max potential of the system to accelerate will be found at some % over no slip is how im understanding the graph. Proven by the information on the braking graph as well.
Also, i have a feeling that the most amount of Gs might be the instant the clutch is let out, and that could be at well over max torque, even max HP, when the kinetic energy is dumped to the ground via the tires. That first instant of launch probably exceeds all other acceleration rates. remember also that even if power was constant, acceleration and rear wheel torque goes down in proportion to speed.
mk
I think a better analogy might be paddles in the water. sure, rack and pinion with a paddle hitting the rack and pushing, would be ideal. But, the faster the paddle moves, the more water it displaces and the greater the force. The friction of tires on the ground as well as the slip % acts in a viscous kind of way. No slip, and you dont maximize your acceleration potential at any speed.
If this is true, then theoretically, you migh be able to generate more force at a slightly higher rpm than at max torque rpm. In the end, this is due to having utilized more available HP. Its one of those things that I know the reason for, but trying to understand how it fits the problem. We all know, at any speed Power determines acceleration and the rear wheel torque. Anyone else want to give it a shot? Its the same reason why airplanes dont accelerate fastest at max torque. max HP always, absorbed by the propeller, provides the greatest thrust. (not max torque of the airplane's engine). if we are talking a motorboat, same thing. However, its much different with tires and the road, due to the slip of the rubber tire. utilizing max potential of the system to accelerate will be found at some % over no slip is how im understanding the graph. Proven by the information on the braking graph as well.
Also, i have a feeling that the most amount of Gs might be the instant the clutch is let out, and that could be at well over max torque, even max HP, when the kinetic energy is dumped to the ground via the tires. That first instant of launch probably exceeds all other acceleration rates. remember also that even if power was constant, acceleration and rear wheel torque goes down in proportion to speed.
mk
I am 100% with you, great explanations. Maximum acceleration is attained in 1st gear if you keep the engine as close to the torque peak for as long as possible. In all other gears, maximum HP is the measure of acceleration. Now, what about "torque multiplication" on an automatic gearbox... ?
Dan
'91 928GT S/C 475hp/460lb.ft
Dan
'91 928GT S/C
475hp/460lb.ft
02-19-2009, 08:50 PM
#68
Rennlist Member
Thread Starter
I, by no means, am touting to know the answer absolutely regarding slip.
This might be a good link that contains the discussion that might provide more info. my only point is that what ever the HP at max torque is, a little faster will provide better acceleration and more torque at the rear wheels due to utilizing more of the available HP. However, the max torque of the engine, if reached at a speed slower than synchronous (i.e. slip) will be the fastest rate of acceleration in that gear. Im not double talking here, Im trying to incorporate how he HP, engine torque and slip are related.
http://groups.google.im/group/rec.au...0f66bf66f2c679
This might be a good link that contains the discussion that might provide more info. my only point is that what ever the HP at max torque is, a little faster will provide better acceleration and more torque at the rear wheels due to utilizing more of the available HP. However, the max torque of the engine, if reached at a speed slower than synchronous (i.e. slip) will be the fastest rate of acceleration in that gear. Im not double talking here, Im trying to incorporate how he HP, engine torque and slip are related.
http://groups.google.im/group/rec.au...0f66bf66f2c679
02-20-2009, 01:42 AM
#69
Rennlist Member
I agree when looking at the graph driving force peaks at we'll say ~20 % slip so I can see how you would interpret this as saying (as long as you are on the up hill slope) increasing slip will increase force put this is actually reading the graph backwards. This graph shows the relation between the maximum drivine force the tire can hold and the wheel slip% generated at that % of max thrust.
So again assuming you are still operating on the up hill side of the curve... as you increase the amount of torque at the rear wheels (driving force) you will move up the curve closer to the peak which represents max thrust capacity of the tire.
Or the other way of rearing the graph is that if you know slip % you can use the graph to determine the % of usable grip you are using. if slip % = 5% you are using ~25% of available traction. If slip % = 20% you are using close to 100% of the available traction. If slip % = 50 % you have over shot the limit and are only getting about 75% of the available thrust.
"The way I am interpreting the graph, (aside from the 0 slip% = 0% driving force) is that whatever the potential power avalable, you will only get a percentage of the acceleration if you dont get a little slip. This little slip, provides the ability of the vehicle to utilize more of the available HP, up to 100% of max power. Remember, Force = power x velocity."
the other problem lies in using Force=Power/velocity. the problem here is that we already established that during clutch slip torque is transmitted on a one to one basis through the clutch but that there is a power loss in the slipping clutch system so engine power does not equal rear wheel power. instead you will need to use F=MA
So again assuming you are still operating on the up hill side of the curve... as you increase the amount of torque at the rear wheels (driving force) you will move up the curve closer to the peak which represents max thrust capacity of the tire.
Or the other way of rearing the graph is that if you know slip % you can use the graph to determine the % of usable grip you are using. if slip % = 5% you are using ~25% of available traction. If slip % = 20% you are using close to 100% of the available traction. If slip % = 50 % you have over shot the limit and are only getting about 75% of the available thrust.
"The way I am interpreting the graph, (aside from the 0 slip% = 0% driving force) is that whatever the potential power avalable, you will only get a percentage of the acceleration if you dont get a little slip. This little slip, provides the ability of the vehicle to utilize more of the available HP, up to 100% of max power. Remember, Force = power x velocity."
the other problem lies in using Force=Power/velocity. the problem here is that we already established that during clutch slip torque is transmitted on a one to one basis through the clutch but that there is a power loss in the slipping clutch system so engine power does not equal rear wheel power. instead you will need to use F=MA
02-20-2009, 02:51 PM
#70
Rennlist Member
Thread Starter
Why is there NO force "0" at " 0" slip on the graph??
I do think you get it in the last part of the 1st paragraph. The graph shows the maximum drving force of the tire, not the max torque or force generate by the engine. Or maybe, if the graph is going to be right, then maybe the driving force is the total force of the engine an what can be generated at the tire. This sounds better, as it would answer the "0" force at "0" slip issue. However, it doesnt answer the "0" slip, (some high percetage of force) issue. Maybe its only a chart of the slipping force, indepedent of the non-slip force. This could be the "slipping force " curve of only the tires. You reach 100% of this, at 15% slip. no slipping force at 0 slip, but the engine and friction of no slip would be independent of that. Now, this makes more sense. thoughts?
Also, think of this for a second. If you were at 35mph as you say, at max torque of the the engine, and 100ft-lbs of torque at the rear wheel. If the slip force can generate more force at any speed, it would be logical to assume that more HP is being utilized at 15-20% slip. This means, more torque could be applied to the rear wheels beyond just the engine torque x gear ratio at max torque rpm.
Think of it this way....... Example: two cars same everything. (engine, gears , tires, etc). both pass through some speed where there is no slip on one car, say, 60mph. This one car is at 4000rpm at max torque, and the other is passing through the same speed at 4400rpm with 10% slip. which one will have a greater acceleration rate?
Right, the slipping car will be traveling at a faster acceleration rate through that same speed. This also means, its utilizing more HP. If it is, then power=fs, then it also will be applying more force to the rear wheels. If you look at that link i provided, you can see how the tires are actually allowing more force to be applied to the pavement, even though the engine torque is less. i.e. at a 10% higher rpm and a higher HP level.
If what you say is true, dragsters would run at max torque out of the hole and leave it there until the tires hooked up. This is not the case. they operate at some level near max hp and try to keep the engine at near max HP while operating at near optimum slip.
as a side note, Im not talking about clutch slip any more as you reference in your last sentence. yes, i agree that clutch slip is just an application of HP loss. Think about if you had a tire that had a 20% optimum slip. you dump the clutch an hook up at 4000rpm at max torque and I dump the clutch and keep my foot in it so when you hit 4000rpm, im at 6000rpm at 20% slip. Im going to run right by you at a faster acceleration rate. Any drag guy will confirm this. why??? power!!
And, ironically, there will be more force transmitted to the pavement as well, even at a lesser engine torque!
mk
I do think you get it in the last part of the 1st paragraph. The graph shows the maximum drving force of the tire, not the max torque or force generate by the engine. Or maybe, if the graph is going to be right, then maybe the driving force is the total force of the engine an what can be generated at the tire. This sounds better, as it would answer the "0" force at "0" slip issue. However, it doesnt answer the "0" slip, (some high percetage of force) issue. Maybe its only a chart of the slipping force, indepedent of the non-slip force. This could be the "slipping force " curve of only the tires. You reach 100% of this, at 15% slip. no slipping force at 0 slip, but the engine and friction of no slip would be independent of that. Now, this makes more sense. thoughts?
Also, think of this for a second. If you were at 35mph as you say, at max torque of the the engine, and 100ft-lbs of torque at the rear wheel. If the slip force can generate more force at any speed, it would be logical to assume that more HP is being utilized at 15-20% slip. This means, more torque could be applied to the rear wheels beyond just the engine torque x gear ratio at max torque rpm.
Think of it this way....... Example: two cars same everything. (engine, gears , tires, etc). both pass through some speed where there is no slip on one car, say, 60mph. This one car is at 4000rpm at max torque, and the other is passing through the same speed at 4400rpm with 10% slip. which one will have a greater acceleration rate?
Right, the slipping car will be traveling at a faster acceleration rate through that same speed. This also means, its utilizing more HP. If it is, then power=fs, then it also will be applying more force to the rear wheels. If you look at that link i provided, you can see how the tires are actually allowing more force to be applied to the pavement, even though the engine torque is less. i.e. at a 10% higher rpm and a higher HP level.
If what you say is true, dragsters would run at max torque out of the hole and leave it there until the tires hooked up. This is not the case. they operate at some level near max hp and try to keep the engine at near max HP while operating at near optimum slip.
as a side note, Im not talking about clutch slip any more as you reference in your last sentence. yes, i agree that clutch slip is just an application of HP loss. Think about if you had a tire that had a 20% optimum slip. you dump the clutch an hook up at 4000rpm at max torque and I dump the clutch and keep my foot in it so when you hit 4000rpm, im at 6000rpm at 20% slip. Im going to run right by you at a faster acceleration rate. Any drag guy will confirm this. why??? power!!
And, ironically, there will be more force transmitted to the pavement as well, even at a lesser engine torque!
mk
I agree when looking at the graph driving force peaks at we'll say ~20 % slip so I can see how you would interpret this as saying (as long as you are on the up hill slope) increasing slip will increase force put this is actually reading the graph backwards. This graph shows the relation between the maximum drivine force the tire can hold and the wheel slip% generated at that % of max thrust.
So again assuming you are still operating on the up hill side of the curve... as you increase the amount of torque at the rear wheels (driving force) you will move up the curve closer to the peak which represents max thrust capacity of the tire.
Or the other way of rearing the graph is that if you know slip % you can use the graph to determine the % of usable grip you are using. if slip % = 5% you are using ~25% of available traction. If slip % = 20% you are using close to 100% of the available traction. If slip % = 50 % you have over shot the limit and are only getting about 75% of the available thrust.
"The way I am interpreting the graph, (aside from the 0 slip% = 0% driving force) is that whatever the potential power avalable, you will only get a percentage of the acceleration if you dont get a little slip. This little slip, provides the ability of the vehicle to utilize more of the available HP, up to 100% of max power. Remember, Force = power x velocity."
the other problem lies in using Force=Power/velocity. the problem here is that we already established that during clutch slip torque is transmitted on a one to one basis through the clutch but that there is a power loss in the slipping clutch system so engine power does not equal rear wheel power. instead you will need to use F=MA
So again assuming you are still operating on the up hill side of the curve... as you increase the amount of torque at the rear wheels (driving force) you will move up the curve closer to the peak which represents max thrust capacity of the tire.
Or the other way of rearing the graph is that if you know slip % you can use the graph to determine the % of usable grip you are using. if slip % = 5% you are using ~25% of available traction. If slip % = 20% you are using close to 100% of the available traction. If slip % = 50 % you have over shot the limit and are only getting about 75% of the available thrust.
"The way I am interpreting the graph, (aside from the 0 slip% = 0% driving force) is that whatever the potential power avalable, you will only get a percentage of the acceleration if you dont get a little slip. This little slip, provides the ability of the vehicle to utilize more of the available HP, up to 100% of max power. Remember, Force = power x velocity."
the other problem lies in using Force=Power/velocity. the problem here is that we already established that during clutch slip torque is transmitted on a one to one basis through the clutch but that there is a power loss in the slipping clutch system so engine power does not equal rear wheel power. instead you will need to use F=MA
Last edited by mark kibort; 02-20-2009 at 07:58 PM.
02-20-2009, 03:24 PM
#71
Rennlist Member
Thread Starter
Slow motion example of slip!
http://www.youtube.com/watch?v=V3yj_OGezWc
They dump the clutch at near 8000rpm,so obviously, peak torque is not a desired operational rpm range.
If you read that link I provided, they discuss contact patch in painful detail. The essence of the discussion talks about slip but the tire tread is distorting. Its distorting but some parts of the tires are still in constant contact with the pavement, even at 15% slip. the entire tire doesnt totally slip unit over 100% slip. In the end, this is a way for the car to take better advantage of the availble HP and has little to do with engine torque values, similarly to what happens at other speeds beyond 1st gear.
http://www.youtube.com/watch?v=V3yj_OGezWc
They dump the clutch at near 8000rpm,so obviously, peak torque is not a desired operational rpm range.
If you read that link I provided, they discuss contact patch in painful detail. The essence of the discussion talks about slip but the tire tread is distorting. Its distorting but some parts of the tires are still in constant contact with the pavement, even at 15% slip. the entire tire doesnt totally slip unit over 100% slip. In the end, this is a way for the car to take better advantage of the availble HP and has little to do with engine torque values, similarly to what happens at other speeds beyond 1st gear.
Last edited by mark kibort; 02-20-2009 at 03:42 PM.
02-21-2009, 02:42 PM
#72
Rennlist Member
Thread Starter
In reviewing that "slip" discussion, I think i may have found a reason for why the slip causes the engine to run at a slightly higher HP range thus producing more rear wheel torque than it would at max torque rpm and speed.
The reason it seems is, due to tire and tread deformation, the tire diameter is effectively smaller, thus giving the 15% greater torque, if the slip is 15%. This also means that the engine is running at 15% higher rpm producing more HP and utilizing more HP as shown by the fact that the tires can apply more force with slip than without slip.
The reason that the tire is effectively smaller in diameter, is not due to it twisting so much that it actually becomes smaller, but due to the tread deformation in front and in the rear of the contact patch where what happens is much of the tread is still in direct contact with the pavement, and other parts are not. this becomes a gear reducer alowing more effective use of the available hp at launch speeds.
Its really the only way that it can be explained, as i do agree, the max torque that would ever be at the rear CV joints would be that found at the engine's max torque in 1st gear. (besides some quck jolts off the line at clutch drop but thats instantaneous force). However, if there can be 15 % slip of the tires, not the clutch, this would provide better acceleration at even the max engine torque rpm and speed, due to an effective gear reduction at the actual surface of the tire.
Chew on that ! Look forward to your comments on this slightly different perspective of Hp and torque at any vehicle speed.
mk
The reason it seems is, due to tire and tread deformation, the tire diameter is effectively smaller, thus giving the 15% greater torque, if the slip is 15%. This also means that the engine is running at 15% higher rpm producing more HP and utilizing more HP as shown by the fact that the tires can apply more force with slip than without slip.
The reason that the tire is effectively smaller in diameter, is not due to it twisting so much that it actually becomes smaller, but due to the tread deformation in front and in the rear of the contact patch where what happens is much of the tread is still in direct contact with the pavement, and other parts are not. this becomes a gear reducer alowing more effective use of the available hp at launch speeds.
Its really the only way that it can be explained, as i do agree, the max torque that would ever be at the rear CV joints would be that found at the engine's max torque in 1st gear. (besides some quck jolts off the line at clutch drop but thats instantaneous force). However, if there can be 15 % slip of the tires, not the clutch, this would provide better acceleration at even the max engine torque rpm and speed, due to an effective gear reduction at the actual surface of the tire.
Chew on that ! Look forward to your comments on this slightly different perspective of Hp and torque at any vehicle speed.
mk
02-21-2009, 09:27 PM
#73
Rennlist Member
Sorry it has been a bit and I haven’t the opportunity to address all your points. I have started to address some in the following but am limited o time so you’ll have to accept this for a starters and I can pick up and late with more comments and try to address any additional points we come up with.
I agree that an engine at high RPM does have a significant amount of inertia that can be used at launch, but I feel this argument is not tailored towards that. I believe the argument was based on torque VS HP generated by the engine on inertia, or you could argue that if you have enough fly wheel you could bring it up to rpm then turn off the engine at launch and still get great acceleration.
OK I think the problem lies in what you believe the left axis of the graph represents. I think you are looking at the thrust produced by the car at peak performance to be the 100% level. The graph is actually independent from the car it is merely tied to a specific tires performance. 100% represents the max thrust capacity of that tire. you could take this tire and install it on a Yugo or a 928 and it will not change the graph or that tire's thrust capacity. The slip angle is related to the % of this thrust capacity you are using that is why 0% slip = 0%force there is no thrust load on the tire and therefore you will get no % slip.
"Why is there NO force "0" at " 0" slip on the graph??
Think of it this way....... Example: two cars same everything. (engine, gears , tires, etc). both pass through some speed where there is no slip on one car, say, 60mph. This one car is at 4000rpm at max torque, and the other is passing through the same speed at 4400rpm with 10% slip. which one will have a greater acceleration rate?
"this is an impossible condition if both cars are running the same tire. it is the thrust load on the tire is what generates the slip angle. If running the same tires by definition the tire subjected to the higher thrust will be operating at the higher slip %.
"I do think you get it in the last part of the 1st paragraph. The graph shows the maximum drving force of the tire, not the max torque or force generate by the engine. Or maybe, if the graph is going to be right, then maybe the driving force is the total force of the engine an what can be generated at the tire. " The tire does not generate the thrust it merely transfers the thrust generated in the form of (torque/radius of tire) to the ground. Therefore the highest thrust that can be transmitted by the tire is at max torque.
" This sounds better, as it would answer the "0" force at "0" slip issue. However, it doesnt answer the "0" slip, (some high percetage of force) issue. "That is because high force = high % slip
"Maybe its only a chart of the slipping force, indepedent of the non-slip force. This could be the "slipping force " curve of only the tires. You reach 100% of this, at 15% slip. no slipping force at 0 slip, but the engine and friction of no slip would be independent of that. Now, this makes more sense. thoughts? …
Also, think of this for a second. If you were at 35mph as you say, at max torque of the the engine, and 100ft-lbs of torque at the rear wheel. If the slip force can generate more force at any speed, it would be logical to assume that more HP is being utilized at 15-20% slip. This means, more torque could be applied to the rear wheels beyond just the engine torque x gear ratio at max torque rpm.
Think of it this way....... Example: two cars same everything. (engine, gears , tires, etc). both pass through some speed where there is no slip on one car, say, 60mph. This one car is at 4000rpm at max torque, and the other is passing through the same speed at 4400rpm with 10% slip. which one will have a greater acceleration rate? "This is impossible as the car at 4000rpm would be generating more than 10% slip as it has a higher torque (thrust load) at the rear tires. In all these scenarios we must assume that the tire has the grip capacity to handle the thrust load generated by the vehicle with out falling off the back side of the slip curve and “loosing” traction by spinning excessively. I think it may help if you are to envision a single car starting from say a 5 mph roll out and then go to full throttle. Initially the torque and hp are both relatively low as engine RPMs are low and in turn thrust at the rear tire is low producing a low slip ratio. Now as you increase in speed and RPM both torque and HP increase, thrust increases as does slip. As you reach max torque of the engine the torque at the rear wheel will also peak which when transferring this torque to the ground through the tire you will generate even more % slip. And as you pass the peak torque and head towards max HP, since the torque has dropped off there is less thrust to be transferred to the ground using a smaller percentage of the tires thrust capacity and therefore slip will also decrease.
"Right, the slipping car will be traveling at a faster acceleration rate through that same speed. This also means, its utilizing more HP. If it is, then power=fs, then it also will be applying more force to the rear wheels. If you look at that link i provided, you can see how the tires are actually allowing more force to be applied to the pavement, even though the engine torque is less. i.e. at a 10% higher rpm and a higher HP level.
If what you say is true, dragsters would run at max torque out of the hole and leave it there until the tires hooked up. This is not the case. they operate at some level near max hp and try to keep the engine at near max HP while operating at near optimum slip.
as a side note, Im not talking about clutch slip any more as you reference in your last sentence. yes, i agree that clutch slip is just an application of HP loss. Think about if you had a tire that had a 20% optimum slip. you dump the clutch an hook up at 4000rpm at max torque and I dump the clutch and keep my foot in it so when you hit 4000rpm, im at 6000rpm at 20% slip. Im going to run right by you at a faster acceleration rate. Any drag guy will confirm this. why??? power!!
And, ironically, there will be more force transmitted to the pavement as well, even at a lesser engine torque!
………………………
In reviewing that "slip" discussion, I think i may have found a reason for why the slip causes the engine to run at a slightly higher HP range thus producing more rear wheel torque than it would at max torque rpm and speed.
"I think you understand what you are trying to say in conjunction with the following argument, but stated it slightly incorrectly. You would have had to say increase thrust not torque as torque at the wheel is independent to tire diameter.
"The reason it seems is, due to tire and tread deformation, the tire diameter is effectively smaller, thus giving the 15% greater torque, if the slip is 15%. This also means that the engine is running at 15% higher rpm producing more HP and utilizing more HP as shown by the fact that the tires can apply more force with slip than without slip.
"Again slip does not increase torque.
I think what your trying to state is that torque over a small lever arm = high force.
"The reason that the tire is effectively smaller in diameter, is not due to it twisting so much that it actually becomes smaller, but due to the tread deformation in front and in the rear of the contact patch where what happens is much of the tread is still in direct contact with the pavement, and other parts are not. this becomes a gear reducer alowing more effective use of the available hp at launch speeds.
"I agree the effective tire diameter is reduces due to side wall deflection.
"Its really the only way that it can be explained, as i do agree, the max torque that would ever be at the rear CV joints would be that found at the engine's max torque in 1st gear. (besides some quck jolts off the line at clutch drop but thats instantaneous force). However, if there can be 15 % slip of the tires, not the clutch, this would provide better acceleration at even the max engine torque rpm and speed, due to an effective gear reduction at the actual surface of the tire.
Chew on that ! Look forward to your comments on this slightly different perspective of Hp and torque at any vehicle speed.
Mk "
I agree that on the reduced tire diameter being able to increase force at the road surface, but this phenomenon occurs at both max HP and max torque. And since you will have more torque at max torque than at max HP to cause tire deflection it can only be that you will get at least as much diameter reduction there and hence an equally proportional increase in thrust due to tire diameter reduction.
Out of time I’ll have to take this up again later.
Simon
I agree that an engine at high RPM does have a significant amount of inertia that can be used at launch, but I feel this argument is not tailored towards that. I believe the argument was based on torque VS HP generated by the engine on inertia, or you could argue that if you have enough fly wheel you could bring it up to rpm then turn off the engine at launch and still get great acceleration.
OK I think the problem lies in what you believe the left axis of the graph represents. I think you are looking at the thrust produced by the car at peak performance to be the 100% level. The graph is actually independent from the car it is merely tied to a specific tires performance. 100% represents the max thrust capacity of that tire. you could take this tire and install it on a Yugo or a 928 and it will not change the graph or that tire's thrust capacity. The slip angle is related to the % of this thrust capacity you are using that is why 0% slip = 0%force there is no thrust load on the tire and therefore you will get no % slip.
"Why is there NO force "0" at " 0" slip on the graph??
Think of it this way....... Example: two cars same everything. (engine, gears , tires, etc). both pass through some speed where there is no slip on one car, say, 60mph. This one car is at 4000rpm at max torque, and the other is passing through the same speed at 4400rpm with 10% slip. which one will have a greater acceleration rate?
"this is an impossible condition if both cars are running the same tire. it is the thrust load on the tire is what generates the slip angle. If running the same tires by definition the tire subjected to the higher thrust will be operating at the higher slip %.
"I do think you get it in the last part of the 1st paragraph. The graph shows the maximum drving force of the tire, not the max torque or force generate by the engine. Or maybe, if the graph is going to be right, then maybe the driving force is the total force of the engine an what can be generated at the tire. " The tire does not generate the thrust it merely transfers the thrust generated in the form of (torque/radius of tire) to the ground. Therefore the highest thrust that can be transmitted by the tire is at max torque.
" This sounds better, as it would answer the "0" force at "0" slip issue. However, it doesnt answer the "0" slip, (some high percetage of force) issue. "That is because high force = high % slip
"Maybe its only a chart of the slipping force, indepedent of the non-slip force. This could be the "slipping force " curve of only the tires. You reach 100% of this, at 15% slip. no slipping force at 0 slip, but the engine and friction of no slip would be independent of that. Now, this makes more sense. thoughts? …
Also, think of this for a second. If you were at 35mph as you say, at max torque of the the engine, and 100ft-lbs of torque at the rear wheel. If the slip force can generate more force at any speed, it would be logical to assume that more HP is being utilized at 15-20% slip. This means, more torque could be applied to the rear wheels beyond just the engine torque x gear ratio at max torque rpm.
Think of it this way....... Example: two cars same everything. (engine, gears , tires, etc). both pass through some speed where there is no slip on one car, say, 60mph. This one car is at 4000rpm at max torque, and the other is passing through the same speed at 4400rpm with 10% slip. which one will have a greater acceleration rate? "This is impossible as the car at 4000rpm would be generating more than 10% slip as it has a higher torque (thrust load) at the rear tires. In all these scenarios we must assume that the tire has the grip capacity to handle the thrust load generated by the vehicle with out falling off the back side of the slip curve and “loosing” traction by spinning excessively. I think it may help if you are to envision a single car starting from say a 5 mph roll out and then go to full throttle. Initially the torque and hp are both relatively low as engine RPMs are low and in turn thrust at the rear tire is low producing a low slip ratio. Now as you increase in speed and RPM both torque and HP increase, thrust increases as does slip. As you reach max torque of the engine the torque at the rear wheel will also peak which when transferring this torque to the ground through the tire you will generate even more % slip. And as you pass the peak torque and head towards max HP, since the torque has dropped off there is less thrust to be transferred to the ground using a smaller percentage of the tires thrust capacity and therefore slip will also decrease.
"Right, the slipping car will be traveling at a faster acceleration rate through that same speed. This also means, its utilizing more HP. If it is, then power=fs, then it also will be applying more force to the rear wheels. If you look at that link i provided, you can see how the tires are actually allowing more force to be applied to the pavement, even though the engine torque is less. i.e. at a 10% higher rpm and a higher HP level.
If what you say is true, dragsters would run at max torque out of the hole and leave it there until the tires hooked up. This is not the case. they operate at some level near max hp and try to keep the engine at near max HP while operating at near optimum slip.
as a side note, Im not talking about clutch slip any more as you reference in your last sentence. yes, i agree that clutch slip is just an application of HP loss. Think about if you had a tire that had a 20% optimum slip. you dump the clutch an hook up at 4000rpm at max torque and I dump the clutch and keep my foot in it so when you hit 4000rpm, im at 6000rpm at 20% slip. Im going to run right by you at a faster acceleration rate. Any drag guy will confirm this. why??? power!!
And, ironically, there will be more force transmitted to the pavement as well, even at a lesser engine torque!
………………………
In reviewing that "slip" discussion, I think i may have found a reason for why the slip causes the engine to run at a slightly higher HP range thus producing more rear wheel torque than it would at max torque rpm and speed.
"I think you understand what you are trying to say in conjunction with the following argument, but stated it slightly incorrectly. You would have had to say increase thrust not torque as torque at the wheel is independent to tire diameter.
"The reason it seems is, due to tire and tread deformation, the tire diameter is effectively smaller, thus giving the 15% greater torque, if the slip is 15%. This also means that the engine is running at 15% higher rpm producing more HP and utilizing more HP as shown by the fact that the tires can apply more force with slip than without slip.
"Again slip does not increase torque.
I think what your trying to state is that torque over a small lever arm = high force.
"The reason that the tire is effectively smaller in diameter, is not due to it twisting so much that it actually becomes smaller, but due to the tread deformation in front and in the rear of the contact patch where what happens is much of the tread is still in direct contact with the pavement, and other parts are not. this becomes a gear reducer alowing more effective use of the available hp at launch speeds.
"I agree the effective tire diameter is reduces due to side wall deflection.
"Its really the only way that it can be explained, as i do agree, the max torque that would ever be at the rear CV joints would be that found at the engine's max torque in 1st gear. (besides some quck jolts off the line at clutch drop but thats instantaneous force). However, if there can be 15 % slip of the tires, not the clutch, this would provide better acceleration at even the max engine torque rpm and speed, due to an effective gear reduction at the actual surface of the tire.
Chew on that ! Look forward to your comments on this slightly different perspective of Hp and torque at any vehicle speed.
Mk "
I agree that on the reduced tire diameter being able to increase force at the road surface, but this phenomenon occurs at both max HP and max torque. And since you will have more torque at max torque than at max HP to cause tire deflection it can only be that you will get at least as much diameter reduction there and hence an equally proportional increase in thrust due to tire diameter reduction.
Out of time I’ll have to take this up again later.
Simon
02-22-2009, 06:18 AM
#74
Rennlist Member
Thread Starter
Yes, you are right, I ment to say, the force would increase at the pavement, not the torque. that is fixed and will follow engine torque as you say.
(i had said that the effective smaller diameter tire, based on slip %, would increase torque and rear wheel forces, but its only the rear wheel forces that would increase due to the shorter lever arm.)
You are also right that this 100% force utilization by the tire, at some optimum slip angle (say 15%) would also be greatest at max torque rpm (say its 4000rpm), even though this might happen at 15% slower vehicle speed if there was 15% slip. Now look at the power being utilized. Lets stay with in your simple gear ratio of 1:1, 100ft-lbs of max torque. we are now at a vehicle speed of what would be found at 3400rpm, but the engine is at 4000rpm. the torque at the rear wheels is 100ft-lbs, but the force due to slip is 15% higher. You are now at the maximum force found at that speed. acceleration will be that speed x that thrust force. power will be 100ft-lbs x 4000rpm/5250 or 76HP.
Thrust force would be 117lbs
Now, look at just changing the slip percentage to 20% and and see what happens at the same vehicle speeed. the engine would be at 4250rpm, call it 95ft-lbs (less engine torque) and the power available would be: 77HP. thrust force would be 118lbs. more power and more thrust force was available, even though the vehicle speed has not changed, AND more thrust force is available in either case vs no slip as that would only utilize lLess than 100lbs of force at that same vehicle speed.
In the two vehicle comparisons, think of one car dumping the clutch and bogging the engine, unable to get slip. the other car getting massive wheel spin. as they pass through any same speed, if one is slipping by x%, it will be utilizing more HP and producing more thrust force. Of course we are talking hypothetically.
so, you are right, max engine torque will determine the max wheel torque in 1st gear, but, there are other variables that will determine the thrust force and the HP utilization at any given vehicle speed.
on the same token, if we look at a vehicle traveling at the peak torque , 4000rpm vehicle speed with no slip (100ft-lbs, 100lbs of force), at slip, of 20%, torque might only be 90ft-lbs, but thrust would be 112lbs and power available would be 87HP. (but we are talking about a higher vehicle speed now) Power dictates the ability to accelerate and the thrust forces generated.
All im doing is trying to fit the law to model. I know for a fact that acceleration is proportional and dictated by power at any vehicle speed. If I can up the rpm during an acceleration while at max torque rpm, i will have more HP and this will result in more accelerative forces (thrust force) This we all know. Now, how to we apply this to the example? We obviously are saying much of the same thing but from different perspectives.
mk
I agree that on the reduced tire diameter being able to increase force at the road surface, but this phenomenon occurs at both max HP and max torque. And since you will have more torque at max torque than at max HP to cause tire deflection it can only be that you will get at least as much diameter reduction there and hence an equally proportional increase in thrust due to tire diameter reduction.
Out of time I’ll have to take this up again later.
Simon
(i had said that the effective smaller diameter tire, based on slip %, would increase torque and rear wheel forces, but its only the rear wheel forces that would increase due to the shorter lever arm.)
You are also right that this 100% force utilization by the tire, at some optimum slip angle (say 15%) would also be greatest at max torque rpm (say its 4000rpm), even though this might happen at 15% slower vehicle speed if there was 15% slip. Now look at the power being utilized. Lets stay with in your simple gear ratio of 1:1, 100ft-lbs of max torque. we are now at a vehicle speed of what would be found at 3400rpm, but the engine is at 4000rpm. the torque at the rear wheels is 100ft-lbs, but the force due to slip is 15% higher. You are now at the maximum force found at that speed. acceleration will be that speed x that thrust force. power will be 100ft-lbs x 4000rpm/5250 or 76HP.
Thrust force would be 117lbs
Now, look at just changing the slip percentage to 20% and and see what happens at the same vehicle speeed. the engine would be at 4250rpm, call it 95ft-lbs (less engine torque) and the power available would be: 77HP. thrust force would be 118lbs. more power and more thrust force was available, even though the vehicle speed has not changed, AND more thrust force is available in either case vs no slip as that would only utilize lLess than 100lbs of force at that same vehicle speed.
In the two vehicle comparisons, think of one car dumping the clutch and bogging the engine, unable to get slip. the other car getting massive wheel spin. as they pass through any same speed, if one is slipping by x%, it will be utilizing more HP and producing more thrust force. Of course we are talking hypothetically.
so, you are right, max engine torque will determine the max wheel torque in 1st gear, but, there are other variables that will determine the thrust force and the HP utilization at any given vehicle speed.
on the same token, if we look at a vehicle traveling at the peak torque , 4000rpm vehicle speed with no slip (100ft-lbs, 100lbs of force), at slip, of 20%, torque might only be 90ft-lbs, but thrust would be 112lbs and power available would be 87HP. (but we are talking about a higher vehicle speed now) Power dictates the ability to accelerate and the thrust forces generated.
All im doing is trying to fit the law to model. I know for a fact that acceleration is proportional and dictated by power at any vehicle speed. If I can up the rpm during an acceleration while at max torque rpm, i will have more HP and this will result in more accelerative forces (thrust force) This we all know. Now, how to we apply this to the example? We obviously are saying much of the same thing but from different perspectives.
mk
I agree that on the reduced tire diameter being able to increase force at the road surface, but this phenomenon occurs at both max HP and max torque. And since you will have more torque at max torque than at max HP to cause tire deflection it can only be that you will get at least as much diameter reduction there and hence an equally proportional increase in thrust due to tire diameter reduction.
Out of time I’ll have to take this up again later.
Simon
02-22-2009, 07:42 AM
#75
Three Wheelin'
Mark, we should have a HUGE spring in the back as a oldy mechanical clock to accumulate torque for when we need it at max HP... 
