Lightweight rotors=less torque?!
12-26-04, 09:32 PM
#51
Super Snuggles
Originally Posted by turbojeff
I haven't seen anything that proves me wrong yet.
Ask Damon, Max, Howard, KevinK2, or any other knowledgeable individual on the forum whose opinion you happen to value.
Torque does not move anything. Power moves the car, power in our case is torque x rpm. Plain and simple, PERIOD. Buy a Physics book and read up.
Torque is a twisting force. What do you think axles do? Horsepower is an arbitrary unit of measure used to describe power exerted over a period time. It's still torque doing the work of twisting the axle to turn the wheel to accelerate the car. The rate of acceleration is highest when torque is highest. Plain and simple. PERIOD.
Where did that graph come from?
What are the units? LB? That isn't a measurement of power.
550lb*ft/sec is one HP IIRC.
So take another look at the graph, make the units so they actually reflect POWER, to me it looks like your trying to show the *thrust* unit. That is NOT a correct way to characterize acceleration.
Maybe you should take a night course at the local community college or something. My guess is that you've never cracked a Physics book in your life.
Last edited by jimlab; 12-26-04 at 09:56 PM.
12-26-04, 10:08 PM
#52
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Originally Posted by jimlab
Sorry Jeff, you are.
Have you ever seen a graph like this?
Correction, that is only part of the reason you accelerate much more slowly in 4th vs. 1st. See above.
Wrong again Jeff. A vehicle's rate of acceleration (g-force) follows the engine's torque curve, not the horsepower curve. Maximum acceleration is at torque peak in first gear, not horsepower peak. Do some research.
Horsepower is merely a calculation based on torque and rpm. If we converted torque at the axles to horsepower at the axles, you'd still see the same drop in POWER in higher gears.
Have you ever seen a graph like this?
Correction, that is only part of the reason you accelerate much more slowly in 4th vs. 1st. See above.
Wrong again Jeff. A vehicle's rate of acceleration (g-force) follows the engine's torque curve, not the horsepower curve. Maximum acceleration is at torque peak in first gear, not horsepower peak. Do some research.
Horsepower is merely a calculation based on torque and rpm. If we converted torque at the axles to horsepower at the axles, you'd still see the same drop in POWER in higher gears.
If you had a perfect continuously variable transmission, the fastest acceleration would be obtained by keeping the engine always at the HP peak. The fastest acceleration in any gear may be at the torque peak but if you changed the gear ratio so that the engine was turning at the HP peak at that same road speed, the car would accelerate harder.
If you draw a line from the HP peak of each gear to the HP peak of the next gear you will have a graph showing the fastest possible acceleration of any possible gears you could install in a car.
By changing gear ratios you can change the torque available at any road speed from nearly zero up to a maximum dictated by the ratio that would place maximum HP at that road speed.
If it turns out that the drag at any speed is greater than the torque generated by the ratio that would put max HP at that road speed, then you can't go that fast no matter what gears you use.
ed
12-26-04, 10:35 PM
#53
Sounds good Ed
.
If you look at the peak acceleration you'll notice in 1st gear the car comes very close to accelerating (or getting the same LB) at the same rate at the HP vs Torque peak, losses aren't much different since the velocity didn't change much.
As the car accelerates the velocity climbs as do the losses. So as the engine climbs up the hp/torque curve the losses are GREATER when the engine reaches peak hp vs peak torque.
POWER is what moves things Jim, not torque, you could torque a flywheel nut @ 250lb*ft but not move it at any rpm with that level of torque.
Jim, if you were setting up the belt drive (basically a CVT) on a snowmobile would you set it up at peak torque or peak power? I'd set it up for peak power, since I know power is what moves things.
.If you look at the peak acceleration you'll notice in 1st gear the car comes very close to accelerating (or getting the same LB) at the same rate at the HP vs Torque peak, losses aren't much different since the velocity didn't change much.
As the car accelerates the velocity climbs as do the losses. So as the engine climbs up the hp/torque curve the losses are GREATER when the engine reaches peak hp vs peak torque.
POWER is what moves things Jim, not torque, you could torque a flywheel nut @ 250lb*ft but not move it at any rpm with that level of torque.
Jim, if you were setting up the belt drive (basically a CVT) on a snowmobile would you set it up at peak torque or peak power? I'd set it up for peak power, since I know power is what moves things.
12-26-04, 10:51 PM
#55
Super Snuggles
If what you're trying to say is that fastest acceleration is achieved by maximizing the torque output of the transmission, then you're right. However, the torque curve falls off at higher rpm, so the maximum torque output of the transmission in any gear is at the torque peak... not the horsepower peak.
What "maximizing the torque output of the transmission" really means is that you should shift at the point at which torque in the current gear falls below torque in the next higher gear. You can find the ideal shift point by analyzing the drive wheel torque curves (as shown in the chart I posted). This isn't misleading, since rate of acceleration (in g's) follows the same curves. If you had a device capable of logging g-force, you'd quickly find this out.
For a stock FD, peak torque is 217 lb-ft. @ 5,000 rpm and peak horsepower is 255 @ 6,500 rpm. The ideal shift points for maximum acceleration are 7,940 rpm for 1st gear to 2nd, 7,710 for 2nd to 3rd, 7,570 for 3rd to 4th, and 7,530 for 4th to 5th. Note that not one of them corresponds to the horsepower peak. They correspond to the points at which torque curves cross in the chart that I posted above. If you shifted at peak horsepower, you'd quickly fall behind someone shifting at the proper shift points.
Gear ratios are selected to drop the engine to or near the torque peak after each shift for a reason; to maximize acceleration in each higher gear.
What "maximizing the torque output of the transmission" really means is that you should shift at the point at which torque in the current gear falls below torque in the next higher gear. You can find the ideal shift point by analyzing the drive wheel torque curves (as shown in the chart I posted). This isn't misleading, since rate of acceleration (in g's) follows the same curves. If you had a device capable of logging g-force, you'd quickly find this out.
For a stock FD, peak torque is 217 lb-ft. @ 5,000 rpm and peak horsepower is 255 @ 6,500 rpm. The ideal shift points for maximum acceleration are 7,940 rpm for 1st gear to 2nd, 7,710 for 2nd to 3rd, 7,570 for 3rd to 4th, and 7,530 for 4th to 5th. Note that not one of them corresponds to the horsepower peak. They correspond to the points at which torque curves cross in the chart that I posted above. If you shifted at peak horsepower, you'd quickly fall behind someone shifting at the proper shift points.
Gear ratios are selected to drop the engine to or near the torque peak after each shift for a reason; to maximize acceleration in each higher gear.
12-26-04, 10:52 PM
#56
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Originally Posted by Compilez
Lighter rotors does not mean lighter compression rotors, right?
-Alex
12-26-04, 11:00 PM
#57
Super Snuggles
Originally Posted by turbojeff
POWER is what moves things Jim, not torque, you could torque a flywheel nut @ 250lb*ft but not move it at any rpm with that level of torque.
12-26-04, 11:43 PM
#59
Originally Posted by dgeesaman
>>What are the units? LB? That isn't a measurement of power.
>lb-ft., and yes it is.
I thought lb-ft was a measure of energy. Just like potential energy - one pound, raised one foot, has one ft-lb of energy.
Dave
>lb-ft., and yes it is.
I thought lb-ft was a measure of energy. Just like potential energy - one pound, raised one foot, has one ft-lb of energy.
Dave
Note
Energy is not measured in kilowatts, but in kilowatt hours (kWh). Mixing up the two units is a very common mistake, so you might want to read the next section on power to understand the difference
Where that came from...
http://www.windpower.org/en/stat/unitsene.htm#eunits
12-26-04, 11:53 PM
#60
Super Snuggles
Originally Posted by dgeesaman
I thought lb-ft was a measure of energy. Just like potential energy - one pound, raised one foot, has one ft-lb of energy.
Originally Posted by turbojeff
1kW = 1.359 hp
Note
Energy is not measured in kilowatts, but in kilowatt hours (kWh). Mixing up the two units is a very common mistake, so you might want to read the next section on power to understand the difference
Where that came from...
http://www.windpower.org/en/stat/unitsene.htm#eunits
Note
Energy is not measured in kilowatts, but in kilowatt hours (kWh). Mixing up the two units is a very common mistake, so you might want to read the next section on power to understand the difference
Where that came from...
http://www.windpower.org/en/stat/unitsene.htm#eunits
12-27-04, 12:12 AM
#61
Originally Posted by jimlab
It's a measure of force. 1 pound of pressure on the end of a 1 foot lever is 1 lb-ft of torque, 2 pounds of pressure on the end of a 0.5 foot lever is 1 lb-ft of torque, etc.
Danish Wind Industry Association? That's what you came up with to prove your point? Nice job, Jeff.
Danish Wind Industry Association? That's what you came up with to prove your point? Nice job, Jeff.
1lbf is a measure of force. IIRC 1lbm on earth is the same.
1lb*ft is a measure of torque OR a force applied on a lever arm.
It is all about the units.
12-27-04, 12:17 AM
#62
Originally Posted by TT_Rex_7
Lighter compression rotors?! You mean lower compression rotors? We're talkign about lightweight rotors, if we take material off of the faces or pockets it will take off weight and also result with a lower compression. Thats not where I'm going to take off the weight so these would be lightweight rotors not lower compression rotors.
-Alex
-Alex
12-27-04, 12:24 AM
#63
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Originally Posted by Compilez
Yeah, sorry. I meant Lower compression rotors. Thanks.
-Alex
12-27-04, 12:28 AM
#64
Originally Posted by TT_Rex_7
Well then, no, this isn't what we're talking about lol. Although like I stated, after I've shed some weight and all goes well, I'm going to work on the pockets on each face wich will effect compression, but give a larger displacement.
-Alex
-Alex
Many versions of the 13b have been produced with different compression ratios (dish on the face of the rotor) but they all have the same displacement.
If you work on the face of your rotor you will have different displacement but a lower compression ratio.
Displacement measures the change in volume in the combustion chamber when the eccentric shaft is rotated.
12-27-04, 12:39 AM
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Originally Posted by turbojeff
I don't think you will get more displacement.
Many versions of the 13b have been produced with different compression ratios (dish on the face of the rotor) but they all have the same displacement.
If you work on the face of your rotor you will have different displacement but a lower compression ratio.
Displacement measures the change in volume in the combustion chamber when the eccentric shaft is rotated.
Many versions of the 13b have been produced with different compression ratios (dish on the face of the rotor) but they all have the same displacement.
If you work on the face of your rotor you will have different displacement but a lower compression ratio.
Displacement measures the change in volume in the combustion chamber when the eccentric shaft is rotated.
-Alex
Last edited by TT_Rex_7; 12-27-04 at 12:46 AM.
12-27-04, 12:46 AM
#66
Yep it would create a larger volume for the combustion chamber but that isn't how displacement is calculated, it only calculates how much volume is displaced when the rotor face moves. The combustion chamber volume will be larger but will still move the same amount you won't have any more displacement.
Sorry I can't explain it more clearly.
Sorry I can't explain it more clearly.
12-27-04, 01:03 AM
#67
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Originally Posted by jimlab
Lightening the rotors will not change the amount of torque the engine produces unless material is removed from the rotor face, affecting displacement.
:
:
-Alex
Last edited by TT_Rex_7; 12-27-04 at 01:08 AM.
12-27-04, 01:29 AM
#69
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It would atleast create more power though correct, since its allowing more air to be put in the motor? If machining the face/pocket will not effect displacement then what could be done to create a larger displacement short of changing eccentric shaft offset for obvious reasons?
-Alex
-Alex
12-27-04, 08:23 AM
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Originally Posted by jimlab
If what you're trying to say is that fastest acceleration is achieved by maximizing the torque output of the transmission, then you're right. However, the torque curve falls off at higher rpm, so the maximum torque output of the transmission in any gear is at the torque peak... not the horsepower peak.
What "maximizing the torque output of the transmission" really means is that you should shift at the point at which torque in the current gear falls below torque in the next higher gear. You can find the ideal shift point by analyzing the drive wheel torque curves (as shown in the chart I posted). This isn't misleading, since rate of acceleration (in g's) follows the same curves. If you had a device capable of logging g-force, you'd quickly find this out.
For a stock FD, peak torque is 217 lb-ft. @ 5,000 rpm and peak horsepower is 255 @ 6,500 rpm. The ideal shift points for maximum acceleration are 7,940 rpm for 1st gear to 2nd, 7,710 for 2nd to 3rd, 7,570 for 3rd to 4th, and 7,530 for 4th to 5th. Note that not one of them corresponds to the horsepower peak. They correspond to the points at which torque curves cross in the chart that I posted above. If you shifted at peak horsepower, you'd quickly fall behind someone shifting at the proper shift points.
Gear ratios are selected to drop the engine to or near the torque peak after each shift for a reason; to maximize acceleration in each higher gear.
What "maximizing the torque output of the transmission" really means is that you should shift at the point at which torque in the current gear falls below torque in the next higher gear. You can find the ideal shift point by analyzing the drive wheel torque curves (as shown in the chart I posted). This isn't misleading, since rate of acceleration (in g's) follows the same curves. If you had a device capable of logging g-force, you'd quickly find this out.
For a stock FD, peak torque is 217 lb-ft. @ 5,000 rpm and peak horsepower is 255 @ 6,500 rpm. The ideal shift points for maximum acceleration are 7,940 rpm for 1st gear to 2nd, 7,710 for 2nd to 3rd, 7,570 for 3rd to 4th, and 7,530 for 4th to 5th. Note that not one of them corresponds to the horsepower peak. They correspond to the points at which torque curves cross in the chart that I posted above. If you shifted at peak horsepower, you'd quickly fall behind someone shifting at the proper shift points.
Gear ratios are selected to drop the engine to or near the torque peak after each shift for a reason; to maximize acceleration in each higher gear.
I find HP equations and graphs are more representative of what I am trying to modify when I am working on a car. Clearly jimlab prefers torque equations and graphs.
No matter how much I like HP graphs, I'll never convince a torque person that my way is better. (I have to admit that sometimes I look at torque graphs when nobody is looking - especially when I am choosing cams for my other car)
I sometimes disagree with the conclusions that jimlab derives from torque graphs but I can't actually say he's wrong.
(but I will say that sometimes he's right for the wrong reason)
ed
12-27-04, 08:31 AM
#71
Originally Posted by turbojeff
1kW = 1.359 hp
Note
Energy is not measured in kilowatts, but in kilowatt hours (kWh). Mixing up the two units is a very common mistake, so you might want to read the next section on power to understand the difference
Where that came from...
http://www.windpower.org/en/stat/unitsene.htm#eunits
Note
Energy is not measured in kilowatts, but in kilowatt hours (kWh). Mixing up the two units is a very common mistake, so you might want to read the next section on power to understand the difference
Where that came from...
http://www.windpower.org/en/stat/unitsene.htm#eunits
Which confirms that ft-lb is not a measurement of power. (I know this isn't important in the big scheme of this thread)
Dave
12-27-04, 07:41 PM
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Originally Posted by TT_Rex_7
It would atleast create more power though correct, since its allowing more air to be put in the motor? If machining the face/pocket will not effect displacement then what could be done to create a larger displacement short of changing eccentric shaft offset for obvious reasons?
-Alex
-Alex
Bump!
-Alex
12-27-04, 08:01 PM
#73
Originally Posted by TT_Rex_7
It would atleast create more power though correct, since its allowing more air to be put in the motor? If machining the face/pocket will not effect displacement then what could be done to create a larger displacement short of changing eccentric shaft offset for obvious reasons?
-Alex
-Alex
1) not change displacement
2) change combustion chamber volume and therefore compression ratio
3) reduce rotating mass of the rotor slightly, improving engine responsiveness
4) probably not improve on Mazda's design unless you really knew what you were doing. If machining the pocket had any significant effect, tuners would be doing it regularly.
Displacement in rotaries is fixed for a particular size of rotor. If you want to create a larger displacement, you can increase the rotor/housing size, add a rotor, or widen the rotors. The second one is the only logical option, and is still a big jump.
I think lightening your rotors is a slightly silly idea unless you've done stress analyses or have other design info to be sure you get it right. In other words, it's easier to put your engine back together now with stock rotors and think about lightening them later than it is to try and lighten them yourself and not get it right.
Dave
12-27-04, 08:27 PM
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Originally Posted by dgeesaman
Machining the pocket would
1) not change displacement
2) change combustion chamber volume and therefore compression ratio
3) reduce rotating mass of the rotor slightly, improving engine responsiveness
4) probably not improve on Mazda's design unless you really knew what you were doing. If machining the pocket had any significant effect, tuners would be doing it regularly.
Displacement in rotaries is fixed for a particular size of rotor. If you want to create a larger displacement, you can increase the rotor/housing size, add a rotor, or widen the rotors. The second one is the only logical option, and is still a big jump.
I think lightening your rotors is a slightly silly idea unless you've done stress analyses or have other design info to be sure you get it right. In other words, it's easier to put your engine back together now with stock rotors and think about lightening them later than it is to try and lighten them yourself and not get it right.
Dave
1) not change displacement
2) change combustion chamber volume and therefore compression ratio
3) reduce rotating mass of the rotor slightly, improving engine responsiveness
4) probably not improve on Mazda's design unless you really knew what you were doing. If machining the pocket had any significant effect, tuners would be doing it regularly.
Displacement in rotaries is fixed for a particular size of rotor. If you want to create a larger displacement, you can increase the rotor/housing size, add a rotor, or widen the rotors. The second one is the only logical option, and is still a big jump.
I think lightening your rotors is a slightly silly idea unless you've done stress analyses or have other design info to be sure you get it right. In other words, it's easier to put your engine back together now with stock rotors and think about lightening them later than it is to try and lighten them yourself and not get it right.
Dave
-Alex