turbos cost hp like superchargers? IMHO no.
01-13-2003, 01:28 PM
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Re: turbos cost hp like superchargers? IMHO no. ChrisB, good job on seperating the different points into readable chuncks.
As far as the statement about a supercharger putting a constant load on the crank. I'm not sure this is exactly true.
When the SC spins up and starts to put up some high boost numbers up I would imagine that compressing the air at this higher rate would cause the load to increase. Also the compressor has some sort of friction coeficient. The higher the revolutions the more your have to multiply this drag component. This seems to be the reason why some SC manufacturers are using teflon in their designs.
As far as the statement about a supercharger putting a constant load on the crank. I'm not sure this is exactly true.
When the SC spins up and starts to put up some high boost numbers up I would imagine that compressing the air at this higher rate would cause the load to increase. Also the compressor has some sort of friction coeficient. The higher the revolutions the more your have to multiply this drag component. This seems to be the reason why some SC manufacturers are using teflon in their designs.
01-13-2003, 03:43 PM
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Re: turbos cost hp like superchargers? IMHO no. The V-1 s-trim has a 3.45:1 step-up ratio, the V-2 s-trim has a 3.60:1 step-up ratio.
The step up ratio doesn't have a big effect on the hp required to turn the blower - it's basically part of a tradeoff - the higher the step-up ratio the higher an impeller speed you can run at a given pulley combination - but as step up ratio increases, so does the "load" placed on the belt (increasing belt slippage problems). If you go to a bigger blower pulley to combat this you reduce your impeller speed, thus boost (it's almost always good to run as big of a crank pulley as possible).
On the other hand lower step up ratio's require smaller pulleys to make the same boost, so even though you have less load on the belt you have to go with a smaller pulley to get into a given rpm range.
The step-up ratio is going to primarily be a function of the intended impeller speed range - the higher the speed, the greater the ratio (generally).
chasgiv3 has it right on the loading - the over-riding factor on loading of the blower is going to be boost level - it takes energy to compress the air - and the more you compress it, the more energy it takes.
Chris
The step up ratio doesn't have a big effect on the hp required to turn the blower - it's basically part of a tradeoff - the higher the step-up ratio the higher an impeller speed you can run at a given pulley combination - but as step up ratio increases, so does the "load" placed on the belt (increasing belt slippage problems). If you go to a bigger blower pulley to combat this you reduce your impeller speed, thus boost (it's almost always good to run as big of a crank pulley as possible).
On the other hand lower step up ratio's require smaller pulleys to make the same boost, so even though you have less load on the belt you have to go with a smaller pulley to get into a given rpm range.
The step-up ratio is going to primarily be a function of the intended impeller speed range - the higher the speed, the greater the ratio (generally).
chasgiv3 has it right on the loading - the over-riding factor on loading of the blower is going to be boost level - it takes energy to compress the air - and the more you compress it, the more energy it takes.
Chris
01-13-2003, 07:05 PM
#23
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Re: turbos cost hp like superchargers? IMHO no. Thanks for explaining the gear step up ratios.
There must be a difference in how much effort in HP to spin the supercharger from model to model and manufacturer. How could/should we compare the above blowers? Even tho HP loss via turning the SC isn't the most important thing to note in a setup, I've never heard it addressed before. I'd think certain SC's would be superior to others in this regard. Any way to compare?
There must be a difference in how much effort in HP to spin the supercharger from model to model and manufacturer. How could/should we compare the above blowers? Even tho HP loss via turning the SC isn't the most important thing to note in a setup, I've never heard it addressed before. I'd think certain SC's would be superior to others in this regard. Any way to compare?
01-13-2003, 07:26 PM
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Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">4 degrees of initial timing, before any type of vacuum advance. Remembver , it s a stang so the computer is a bit low-tek compared to ours. At WOT he should pick up the proper amount of advance.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">yes 4 degrees.. the computer adds up to 16 so my max timing was at 20 degrees... and no, vacuum plays no part in it... true the computers are not as advanced as the newer cars are.. though my stangs puter added timing based on load and throttle position.
01-13-2003, 07:45 PM
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Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by ChrisB:
<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Who thinks the rwtq values are real. They arn't. The "torque" we see on dynographs is actually work. (...) Add all 4 together for a value in lbs (lbs is a unit of force) then multiply it by the stroke but convert it to feet. You get WORK (not torque) which is in lbft.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
I disagree with that - you are measuring torque, even in your example. Torque is the rotational analogue of force, and is force * length of the moment arm (1/2 the stroke actually), which is exactly what you describe above. Work is the distance over which force is applied - e.g. I push a box with 5lbs of force over 10 feet.
Think about it - I can apply 10lbs of force to a 1' radius crankshaft and not have it move. I have still applied a torque of 10 ft-lbs, but no work has been done as it hasn't mooved. Or even if it does move then the work would be the force applied * the arc length traveled by the object at the end of the crankshaft (assuming everything else has 0 mass, to make it simple).
If your method is correct then the work done rotating an object 25 times is no different than rotation it 1 time - as long as the mass and radius are the same. (and obviously that isn't correct).
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Max velocity of the piston is directly between TDC and BDC. Max force on the rod is right on TDC or BDC.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
Yep, though exactly where you get max velocity depends on your r/s ratio - here is a spreadsheet I had done awhile ago that graphs the piston position and it's derivatives through a crank revolution (from TDC to TDC) - you can change the first value in the data table heading to try different combinations.
http://www.slowcar.net/shared/pistonmotion.xls
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">This is why turbos or boost is most often not as hard on the engine as high compression NA and or nitrous. Because the inertial load is the slope of the cos function, and a boosted powerstroke produces less peak force then a high compression PS(powerstroke) and follows a less steep decline in force until BDC so it follows the cos curve of the inertial load better. But because the deline in force on a turbocharged PS is less steep, the total work can be greater then the PS of a high compression ratio NA engine because it's the area under the curve that counts.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I don't really accept this statement either, especially the bolded part (emphasis mine). When you talk about the peak forces on the power stroke, and we assume the same piston area, you are basically talking about cylinder pressure, which is directly analgous to torque.
You can make FAR more cylinder pressure/torque with a turbo (or blower setup) for that matter than you can with a NA setup - so how can the FI setup help but have more cylinder pressure and thus force on the power stroke. It seems to me you claim that it is because a smaller force is averaged out over a longer period, but the first part of that is not true - it is a not a smaller force. It will not be twice as much at twice the torque levels, because of the area under the curve effect you mention, but it will be greater. Here is a diagram from corky bell's book
<img src="http://www.slowcar.net/shared/cylprs.gif" alt=" - " />
This clearly shows cylinder pressure (thus force) going up across the board - peak and all.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Oh yeah and superchargers put a constant load on the crank a negative load. </font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
How is the supercharger load a negative load?
Everything else aside, turbo's cost horsepower also - if only from increased pumping losses (having to work against a pressurized exhaust side). Yes they are definitely more efficient, as the derive a good bit of their energy from a source that is otherwise wasted, so you improve the overall effeciency of the motor. But they still take horsepower to turn, and still "cost" some in a theoretical sense (and in that same sense a blower costs more) - but both obviously make net power.
Chris </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">ChrisB I was comparing an NA car with the same hp as a turbo car, maybe the turbo car would only have to run 5 psi to make up for a really high compression radical cam NA car. If you were to overlay the two graphs of same hp cars the turbo car's graph would have a lower peak and then part way down the graph it would cross NA cars graph and have higher force the rest of the way.
I knew someone was going to argue against my point about torque. ok let's say I go get a 2 foot long wrench, and apply it to a nut on one of the tires, and put all my weight on it. That would be about 400 lbfeet of torque, so I guess the car is going to take off and bust a 1.8 60'. Nope. but that's 400 lb feet of RWtorque. The force of the piston over the ditance is the work done by the piston, you can combine work, that is add work up by revolutions/second to get power, but you CANNOT add torques up by revolution/second.
Think about it. A constant no rev based source of torque would put out the most power. If it's pulsing like a motor, then the interruption would cause there to be less power.
But work, a certain amount of work done, is done no matter how long it takes. Thus a rev based source of work would put out more power as the revs/second increase, because 400 lb ft of work at 5000 rpm is like having 5000 dudes standing on wrenches moving the wrench 2 ft with 200 lb of force. That would be 2,000,000 lb ft of total torque for that split second, whereas 400 lb ft of torque at 5000 rpm is like having 1 dude like me standing on a wrench for a split second. Remember people, torque is a rotational force. Work is the process of converting energy. Now go look in a physics book and tell me why work and power are in the same chapter or even the same section, and torque is nowhere to be found.
"Power is the rate of doing work (or rate of conversion of energy)". -- oops did I quote that from a physics text. Rate of doing work huh. Is that why power is a directly proportional to "torque" (work) and rpm? wait isn't the formula for hp "torque" * rpm / some constant or something like that.
wait here's some more physics
Power = work / time. Or better yet Power = (total work) / time. Or better yet POwer = (work done by 8 pistons in one rev * rpm) / 1 min. You need some constant to convert this unit into hp.
booooyaaaaa bizzzzootch!!! <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
hey have a <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" /> it's just debate
<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Who thinks the rwtq values are real. They arn't. The "torque" we see on dynographs is actually work. (...) Add all 4 together for a value in lbs (lbs is a unit of force) then multiply it by the stroke but convert it to feet. You get WORK (not torque) which is in lbft.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
I disagree with that - you are measuring torque, even in your example. Torque is the rotational analogue of force, and is force * length of the moment arm (1/2 the stroke actually), which is exactly what you describe above. Work is the distance over which force is applied - e.g. I push a box with 5lbs of force over 10 feet.
Think about it - I can apply 10lbs of force to a 1' radius crankshaft and not have it move. I have still applied a torque of 10 ft-lbs, but no work has been done as it hasn't mooved. Or even if it does move then the work would be the force applied * the arc length traveled by the object at the end of the crankshaft (assuming everything else has 0 mass, to make it simple).
If your method is correct then the work done rotating an object 25 times is no different than rotation it 1 time - as long as the mass and radius are the same. (and obviously that isn't correct).
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Max velocity of the piston is directly between TDC and BDC. Max force on the rod is right on TDC or BDC.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
Yep, though exactly where you get max velocity depends on your r/s ratio - here is a spreadsheet I had done awhile ago that graphs the piston position and it's derivatives through a crank revolution (from TDC to TDC) - you can change the first value in the data table heading to try different combinations.
http://www.slowcar.net/shared/pistonmotion.xls
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">This is why turbos or boost is most often not as hard on the engine as high compression NA and or nitrous. Because the inertial load is the slope of the cos function, and a boosted powerstroke produces less peak force then a high compression PS(powerstroke) and follows a less steep decline in force until BDC so it follows the cos curve of the inertial load better. But because the deline in force on a turbocharged PS is less steep, the total work can be greater then the PS of a high compression ratio NA engine because it's the area under the curve that counts.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I don't really accept this statement either, especially the bolded part (emphasis mine). When you talk about the peak forces on the power stroke, and we assume the same piston area, you are basically talking about cylinder pressure, which is directly analgous to torque.
You can make FAR more cylinder pressure/torque with a turbo (or blower setup) for that matter than you can with a NA setup - so how can the FI setup help but have more cylinder pressure and thus force on the power stroke. It seems to me you claim that it is because a smaller force is averaged out over a longer period, but the first part of that is not true - it is a not a smaller force. It will not be twice as much at twice the torque levels, because of the area under the curve effect you mention, but it will be greater. Here is a diagram from corky bell's book
<img src="http://www.slowcar.net/shared/cylprs.gif" alt=" - " />
This clearly shows cylinder pressure (thus force) going up across the board - peak and all.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Oh yeah and superchargers put a constant load on the crank a negative load. </font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
How is the supercharger load a negative load?
Everything else aside, turbo's cost horsepower also - if only from increased pumping losses (having to work against a pressurized exhaust side). Yes they are definitely more efficient, as the derive a good bit of their energy from a source that is otherwise wasted, so you improve the overall effeciency of the motor. But they still take horsepower to turn, and still "cost" some in a theoretical sense (and in that same sense a blower costs more) - but both obviously make net power.
Chris </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">ChrisB I was comparing an NA car with the same hp as a turbo car, maybe the turbo car would only have to run 5 psi to make up for a really high compression radical cam NA car. If you were to overlay the two graphs of same hp cars the turbo car's graph would have a lower peak and then part way down the graph it would cross NA cars graph and have higher force the rest of the way.
I knew someone was going to argue against my point about torque. ok let's say I go get a 2 foot long wrench, and apply it to a nut on one of the tires, and put all my weight on it. That would be about 400 lbfeet of torque, so I guess the car is going to take off and bust a 1.8 60'. Nope. but that's 400 lb feet of RWtorque. The force of the piston over the ditance is the work done by the piston, you can combine work, that is add work up by revolutions/second to get power, but you CANNOT add torques up by revolution/second.
Think about it. A constant no rev based source of torque would put out the most power. If it's pulsing like a motor, then the interruption would cause there to be less power.
But work, a certain amount of work done, is done no matter how long it takes. Thus a rev based source of work would put out more power as the revs/second increase, because 400 lb ft of work at 5000 rpm is like having 5000 dudes standing on wrenches moving the wrench 2 ft with 200 lb of force. That would be 2,000,000 lb ft of total torque for that split second, whereas 400 lb ft of torque at 5000 rpm is like having 1 dude like me standing on a wrench for a split second. Remember people, torque is a rotational force. Work is the process of converting energy. Now go look in a physics book and tell me why work and power are in the same chapter or even the same section, and torque is nowhere to be found.
"Power is the rate of doing work (or rate of conversion of energy)". -- oops did I quote that from a physics text. Rate of doing work huh. Is that why power is a directly proportional to "torque" (work) and rpm? wait isn't the formula for hp "torque" * rpm / some constant or something like that.
wait here's some more physics
Power = work / time. Or better yet Power = (total work) / time. Or better yet POwer = (work done by 8 pistons in one rev * rpm) / 1 min. You need some constant to convert this unit into hp.
booooyaaaaa bizzzzootch!!! <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
hey have a <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" /> it's just debate
01-13-2003, 08:05 PM
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Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by chasgiv3:
<strong> ChrisB, good job on seperating the different points into readable chuncks.
As far as the statement about a supercharger putting a constant load on the crank. I'm not sure this is exactly true.
When the SC spins up and starts to put up some high boost numbers up I would imagine that compressing the air at this higher rate would cause the load to increase. Also the compressor has some sort of friction coeficient. The higher the revolutions the more your have to multiply this drag component. This seems to be the reason why some SC manufacturers are using teflon in their designs. </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">true but what I meant was constant as opposed to pulsing with the pistons. The load at 5000 rpm is there at all points of each revolution of the crank. That's what I meant and by negative load I mean the load the supercharger puts on the crank by the belt. not the boost. It is a negative load if it eats up hp.
Hey Chris B, after looking at that graph, it shows my point completely. Look how the turbo one extends force past where the NA one falls off completely, and notice also how the highest point of force (that's the force we worry about) for the turbo one which at 2X atmosphereic will just about double the hp of the NA car.
but notice that the max force on the turbo charged piston on that graph isn't anywhere near double the max force on the NA piston. This was just on of my drunken points. <img border="0" alt="[cheers]" title="" src="graemlins/gr_cheers.gif" />
<strong> ChrisB, good job on seperating the different points into readable chuncks.
As far as the statement about a supercharger putting a constant load on the crank. I'm not sure this is exactly true.
When the SC spins up and starts to put up some high boost numbers up I would imagine that compressing the air at this higher rate would cause the load to increase. Also the compressor has some sort of friction coeficient. The higher the revolutions the more your have to multiply this drag component. This seems to be the reason why some SC manufacturers are using teflon in their designs. </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">true but what I meant was constant as opposed to pulsing with the pistons. The load at 5000 rpm is there at all points of each revolution of the crank. That's what I meant and by negative load I mean the load the supercharger puts on the crank by the belt. not the boost. It is a negative load if it eats up hp.
Hey Chris B, after looking at that graph, it shows my point completely. Look how the turbo one extends force past where the NA one falls off completely, and notice also how the highest point of force (that's the force we worry about) for the turbo one which at 2X atmosphereic will just about double the hp of the NA car.
but notice that the max force on the turbo charged piston on that graph isn't anywhere near double the max force on the NA piston. This was just on of my drunken points. <img border="0" alt="[cheers]" title="" src="graemlins/gr_cheers.gif" />
01-14-2003, 12:53 AM
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Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by MelloYellow:
<strong> 15% .. Interesting.
I think the Prochargers use steeper gears in their drive than the Vortechs FWIW. </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">not all of them...i am not aware of the vortech ratios but only the new F-series procharger blowers have the large step-up gear ratios, i believe its 5:1 or something like that
you can see the differences at www.procharger.com
<strong> 15% .. Interesting.
I think the Prochargers use steeper gears in their drive than the Vortechs FWIW. </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">not all of them...i am not aware of the vortech ratios but only the new F-series procharger blowers have the large step-up gear ratios, i believe its 5:1 or something like that
you can see the differences at www.procharger.com
01-14-2003, 07:19 AM
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Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by AlienDroid:
<strong>ChrisB I was comparing an NA car with the same hp as a turbo car, maybe the turbo car would only have to run 5 psi to make up for a really high compression radical cam NA car. If you were to overlay the two graphs of same hp cars the turbo car's graph would have a lower peak and then part way down the graph it would cross NA cars graph and have higher force the rest of the way.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif"></strong>
1) What is the point of doing a turbo car that only makes the same power as a NA car? Realistically everyone doing a turbo here is doing it to make more power/torque
2) You would have to specify the same power @ the same rpm range - the turbo at the same power level could easily have much more torque/force if it does it at 1/2 the rpm that the NA car does.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">
I knew someone was going to argue against my point about torque. ok let's say I go get a 2 foot long wrench, and apply it to a nut on one of the tires, and put all my weight on it. That would be about 400 lbfeet of torque, so I guess the car is going to take off and bust a 1.8 60'. Nope. but that's 400 lb feet of RWtorque.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
1) I think you are getting work and torque here confused because of the way an engine works - it must be rotating to continue operation (e.g. it can't mimick the "stuck nut" metaphor) - but that doesn't mean torque and work are the same thing - they are still 2 separate entities;
2) Torque has nothing to do with your sixty foot time - all that maters is the power (horsepower) seen over the rpm range you observe when you launch. Higher torque @ an rpm = higher hp, but it is the hp unit you must look at (as rpm isn't fixed)
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">
[...]
</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
Sorry, but I don't really see the point of the rest of your post. You went on to type a simple primer on some basic physics - and that is all well and good, but nothing you typed supported your contention that the ft-lb values we get at a dyno are actually work (which they are not).
Even the formula you provided for power (power = work/time) contradicts this - rpm's aren't a unit of time, they are a unit of radial velocity - so to make the relationship horsepower = torque * rpm / 5250 true we need to use a formula for power that contains velocity. How about power = force * velocity (just a different way of writing your formula above). or written differently,
horsepower (power) = torque (force) * rpm (velocity) * K (proportonality constant, 1/5250)
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Hey Chris B, after looking at that graph, it shows my point completely. Look how the turbo one extends force past where the NA one falls off completely, and notice also how the highest point of force (that's the force we worry about) for the turbo one which at 2X atmosphereic will just about double the hp of the NA car</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
I of course agree with you provided the hp and rpm values of both cars are pretty much identical - it was just you didn't mention that condition previously - and again, realistically, what's the point of building a turbo car that only makes NA level power/torque.
Chris
<strong>ChrisB I was comparing an NA car with the same hp as a turbo car, maybe the turbo car would only have to run 5 psi to make up for a really high compression radical cam NA car. If you were to overlay the two graphs of same hp cars the turbo car's graph would have a lower peak and then part way down the graph it would cross NA cars graph and have higher force the rest of the way.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif"></strong>
1) What is the point of doing a turbo car that only makes the same power as a NA car? Realistically everyone doing a turbo here is doing it to make more power/torque
2) You would have to specify the same power @ the same rpm range - the turbo at the same power level could easily have much more torque/force if it does it at 1/2 the rpm that the NA car does.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">
I knew someone was going to argue against my point about torque. ok let's say I go get a 2 foot long wrench, and apply it to a nut on one of the tires, and put all my weight on it. That would be about 400 lbfeet of torque, so I guess the car is going to take off and bust a 1.8 60'. Nope. but that's 400 lb feet of RWtorque.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
1) I think you are getting work and torque here confused because of the way an engine works - it must be rotating to continue operation (e.g. it can't mimick the "stuck nut" metaphor) - but that doesn't mean torque and work are the same thing - they are still 2 separate entities;
2) Torque has nothing to do with your sixty foot time - all that maters is the power (horsepower) seen over the rpm range you observe when you launch. Higher torque @ an rpm = higher hp, but it is the hp unit you must look at (as rpm isn't fixed)
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">
[...]
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Sorry, but I don't really see the point of the rest of your post. You went on to type a simple primer on some basic physics - and that is all well and good, but nothing you typed supported your contention that the ft-lb values we get at a dyno are actually work (which they are not).
Even the formula you provided for power (power = work/time) contradicts this - rpm's aren't a unit of time, they are a unit of radial velocity - so to make the relationship horsepower = torque * rpm / 5250 true we need to use a formula for power that contains velocity. How about power = force * velocity (just a different way of writing your formula above). or written differently,
horsepower (power) = torque (force) * rpm (velocity) * K (proportonality constant, 1/5250)
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Hey Chris B, after looking at that graph, it shows my point completely. Look how the turbo one extends force past where the NA one falls off completely, and notice also how the highest point of force (that's the force we worry about) for the turbo one which at 2X atmosphereic will just about double the hp of the NA car</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
I of course agree with you provided the hp and rpm values of both cars are pretty much identical - it was just you didn't mention that condition previously - and again, realistically, what's the point of building a turbo car that only makes NA level power/torque.
Chris
01-14-2003, 08:22 PM
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Re: turbos cost hp like superchargers? IMHO no. "I of course agree with you provided the hp and rpm values of both cars are pretty much identical - it was just you didn't mention that condition previously - and again, realistically, what's the point of building a turbo car that only makes NA level power/torque."
the point would be to observe strain on the motor. You can make 450rwhp NA with very high compression on race gas with a very agressive heads/cam setup (go check out the second on engine internals, it's been done)
You can also make 450rwhp on a turbotech t76 kit. That's the same amount of rwhp. So the point is to observe the force on the crank thoughout the power stroke of each at some point where they are making the same amound of "torque". You still don't get it. ok, go get a bike, on each stroke, slam your foot down hard suddenly and let off the rest of the way down. Then try doing a more even stroke, but do it so you accelerate at the same rate. You can see for yourself that the first example is more damaging to the equipment then the second.
THAT IS THE POINT to that.
and btw don't take anything offensive, I'm just trying to debate my points here. <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" /> I'm going to reply one at a time to your reply.
the point would be to observe strain on the motor. You can make 450rwhp NA with very high compression on race gas with a very agressive heads/cam setup (go check out the second on engine internals, it's been done)
You can also make 450rwhp on a turbotech t76 kit. That's the same amount of rwhp. So the point is to observe the force on the crank thoughout the power stroke of each at some point where they are making the same amound of "torque". You still don't get it. ok, go get a bike, on each stroke, slam your foot down hard suddenly and let off the rest of the way down. Then try doing a more even stroke, but do it so you accelerate at the same rate. You can see for yourself that the first example is more damaging to the equipment then the second.
THAT IS THE POINT to that.
and btw don't take anything offensive, I'm just trying to debate my points here. <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" /> I'm going to reply one at a time to your reply.
01-14-2003, 08:58 PM
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Re: turbos cost hp like superchargers? IMHO no. "Sorry, but I don't really see the point of the rest of your post. You went on to type a simple primer on some basic physics - and that is all well and good, but nothing you typed supported your contention that the ft-lb values we get at a dyno are actually work (which they are not).
Even the formula you provided for power (power = work/time) contradicts this - rpm's aren't a unit of time, they are a unit of radial velocity - so to make the relationship horsepower = torque * rpm / 5250 true we need to use a formula for power that contains velocity. How about power = force * velocity (just a different way of writing your formula above). or written differently,
horsepower (power) = torque (force) * rpm (velocity) * K (proportonality constant, 1/5250)"
I think you may not have noticed that the force placed on the piston is a linear force that is converted to a roatational force via the crank. Linear force of distance (stroke) is work. I still consider it work <img border="0" title="" alt="[Razz]" src="gr_images/icons/tongue.gif" />
Even the formula you provided for power (power = work/time) contradicts this - rpm's aren't a unit of time, they are a unit of radial velocity - so to make the relationship horsepower = torque * rpm / 5250 true we need to use a formula for power that contains velocity. How about power = force * velocity (just a different way of writing your formula above). or written differently,
horsepower (power) = torque (force) * rpm (velocity) * K (proportonality constant, 1/5250)"
I think you may not have noticed that the force placed on the piston is a linear force that is converted to a roatational force via the crank. Linear force of distance (stroke) is work. I still consider it work <img border="0" title="" alt="[Razz]" src="gr_images/icons/tongue.gif" />
01-14-2003, 09:20 PM
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Re: turbos cost hp like superchargers? IMHO no. Re NA vs turbo power - I don't accept the contention the motors of the same rwhp have less peak cylinder pressure in turbo form than NA form (same rwhp based on your 450rwhp example).
I understand your point, but there are other variables also. The turbo motor *will* make it's power at a much lower rpm range than the NA motor - to do this it *has* to produce more cylinder pressure - 450rwhp @ 5500 rpm is a result of a higher torque @ that rpm (and thus pressure) than 450rwhp at 6500rpm.
If you stipulate that the power curves are identical with respect to rpm as well as power then I would tend to agree that the peak cylinder pressures may be less - but I don't think that's a valid real world scenario.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by AlienDroid:
<strong>I think you may not have noticed that the force placed on the piston is a linear force that is converted to a roatational force via the crank. Linear force of distance (stroke) is work. I still consider it work <img border="0" title="" alt="[Razz]" src="gr_images/icons/tongue.gif" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Yes, force applied over a distance is work - but in your example:
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Linear force of distance (stroke) is work</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">you are talking about force applied AT a distance (through a moment arm) - not OVER a distance. Force applied to a moment arm is a torque, not work. (besides the fact that the distance it is applied at is actually half the stroke)
If I have a 1 lb weight on a 1 ft moment arm - according to your definition that is analgous to one lb-ft of work. But work needs motion - correct? So let's say the crank turns 1 revolution - that is a linear distance of 2pi feet. So in one revolution my work done is 2pi feet * 1lb , or 2pi lb-ft. The stroke itself is irrelevant - only the linear distance applied. If the half the stroke is the "foot" unit in work as you claim above then it doesn't matter if I do one revolution or 50 - it is still 1 lb at 1 ft - yet obviously more work is done in 50 revolutions than 1.
And again, if your contention that "the ft-lbs measured on a dyno are not torque, but work" is true, how to you explain the relationship of
horsepower = torque * rpm / 5250.
If the torque really is work then the equation states that power = work * velocity (which is incorrect) - thus I also submit that torque is NOT the same as work, and that the values measured on a dyno are NOT work, but rather, torque (as they are represented)
Chris
I understand your point, but there are other variables also. The turbo motor *will* make it's power at a much lower rpm range than the NA motor - to do this it *has* to produce more cylinder pressure - 450rwhp @ 5500 rpm is a result of a higher torque @ that rpm (and thus pressure) than 450rwhp at 6500rpm.
If you stipulate that the power curves are identical with respect to rpm as well as power then I would tend to agree that the peak cylinder pressures may be less - but I don't think that's a valid real world scenario.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by AlienDroid:
<strong>I think you may not have noticed that the force placed on the piston is a linear force that is converted to a roatational force via the crank. Linear force of distance (stroke) is work. I still consider it work <img border="0" title="" alt="[Razz]" src="gr_images/icons/tongue.gif" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Yes, force applied over a distance is work - but in your example:
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Linear force of distance (stroke) is work</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">you are talking about force applied AT a distance (through a moment arm) - not OVER a distance. Force applied to a moment arm is a torque, not work. (besides the fact that the distance it is applied at is actually half the stroke)
If I have a 1 lb weight on a 1 ft moment arm - according to your definition that is analgous to one lb-ft of work. But work needs motion - correct? So let's say the crank turns 1 revolution - that is a linear distance of 2pi feet. So in one revolution my work done is 2pi feet * 1lb , or 2pi lb-ft. The stroke itself is irrelevant - only the linear distance applied. If the half the stroke is the "foot" unit in work as you claim above then it doesn't matter if I do one revolution or 50 - it is still 1 lb at 1 ft - yet obviously more work is done in 50 revolutions than 1.
And again, if your contention that "the ft-lbs measured on a dyno are not torque, but work" is true, how to you explain the relationship of
horsepower = torque * rpm / 5250.
If the torque really is work then the equation states that power = work * velocity (which is incorrect) - thus I also submit that torque is NOT the same as work, and that the values measured on a dyno are NOT work, but rather, torque (as they are represented)
Chris
01-16-2003, 08:29 PM
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Re: turbos cost hp like superchargers? IMHO no. well guys...this is kinda off topic and i'm not as good at physics as you guys but i was wondering about the hp-"torque" equation. Accoding to that, hp and "torque" should always be porportional, and how come i've seen some figure of hp at a certain rpm and when i use the equation i dont get the same torque number that i got from the dyno. it may be a stupid question, but we are all trying to learn here. thanks
01-17-2003, 06:25 AM
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Re: turbos cost hp like superchargers? IMHO no. Make sure you are reading the hp and the torque numbers from the correct axis - quite often one value will be scaled on the left side of the y-axis, while the other will be scaled on the right side (also why the don't neccecarily cross graphically at 5250)
Chris
Chris
