SCCAForums.com

DaveH:

A = 11.71 m/s^2 = 1.2 g

BTW, this is a 2.3 second 0-60 if you can keep the tires stuck (ignoring aero drag).

cmt52663:

an it'll probably pass a sound check huh...

Of course it will...  The EMP that's created as this thing accelerates will blow any electrical device within 100'.  The courseworkers will have to leave their wallets in paddock or their CC magnetic strips will be erased.  I'm sure that whoever creates this thing will create a Gaussian box that protect said wallets/CCs by their support vehicle for anyone who *cough* wants to protect their stuff.
 

DaveH:

Gaussian boxes are for electric fields.  For magnetic fields you need several inches of ferrous material.  Everybody make sure to bring cash to Topeka!

 Shhhhh!!!!  I'm trying to get people to leave their wallets at our truck.     How else are we going to pay for the batteries?

 

Bringing this one back up to the top.  It is good to see the discussion and some educated mathematics added to it.  It is amazing how often the torque and horsepower issues are misunderstood. Hell it's even published misleadingly in the Speed Secrets book where it says "torque wins races, hp sells cars."   The simple fact is that area under the horsepower curve is typically what matters, and your wheel torque is generated by your gearing.  The reason being that gearing is a torque multiplier and therefore can overcome much of the deficiencies of a low torque motor.  Nobody here probably wants to race a formula 1 car in a drag race even though they typically "only" make about ~260 ft-lb's of torque even though they were until the newer restrictions and turbo restrictions often just shy of 1000 hp cars at 20,000 rpm's (1000 hp at 20,000 rpm's is 260 ft-lbs of torque).  This is made available at any given speed based upon your gearing.  Therefore the discussion between torque and horsepower HAS to include gearing.  It also has to include the speed range intended and application desired.  You basically want to have the widest area at your lowest possible gear that fits your desired range. Or a set of gearing if the range exceeds the capabilities of the engine.  Also, another common misconception is that you want to shift earlier than redline.  The ONLY time this would happen is if your engine falls off significantly enough in hp and therefore torque such that the mechanical advantage of the next gear at the rpm it will enter at actually exceeds your current output.   As a simple example, and since I have access to the numbers, a Mazda Protege 2nd gear is 1.842 and 3rd is 1.310 (final drive is the same so no need to include in the caluclations).  So assuming a torque value of 100 ft-lbs at 4,000 rpm's and a torque value of 80 ft-lbs at 7000 rpm's (roughly accurate) 2nd gear is able to create 184 ft-lbs to the final drive at 4,000 and 147 ft lbs to final drive at 7,000. 3rd gear creates 131 ft-lbs at 4,000 and 105 ft-lbs at 7,000.  Now, I didn't account for what the actual rpm set would be as that requires quite a few calculations, but this shows that even though there was a 20% drop in output of the engine going toward redline, the shift up to the next gear has a great enough mechanical advantage loss that even though you are back in the peak range of the engine at the beginning of that shift you have actually LESS torque delivered to the wheels because of the effect of gearing. 

 So again, the area under the curve is your critical factor and the horsepower number is what matters most as the gearing will compensate for the torque values, and when it comes to when to shift your gearing ratios relative to the reflected loss of the output curve of the engine will dominate whether a shift is better, and in most cases on production cars these days redline will serve better.  A lot of discussion can be had though on the effects of short gearing (requires more shifts) etc as well and those are all factors to be reckoned with. 

Now if you hold gearing constant though on a given vehicle, and the speed range equivalent the discussion gets a little more muddled, as you can get into some iffy examples, but simplified again the vehicle with more torque will also have a greater area under the curve for the horsepower and thus will be faster. This is why many people think that torque wins due to the days of the big V8's at lower rpm's... the guys posting higher torque numbers typically had fatter horspower bands as well even though they may not have had high horsepower numbers (especially if redline isn't much past the 5252 rpm mark). 

And on the electric car, remember that magnetic fields decay exponentially relative to distance so you'd have to be pretty close for the fields to be strong enough... definitely stay away with the pacemakers :)

I agree in principle, but one thing I think is often under appreciated is the amount of torque/hp required simply to accelerate the rotating inertia of the engine components, flywheel, clutch, transmission input shaft etc. especially in the lower gears.  In first gear, this might be 20% of the total available engine torque/hp.  If you double the gear ratio, the engine has to accelerate twice as quickly, and therefore requires twice as much torque/hp to accelerate the rotating masses.  Therefore, you don't get twice as much torque at the wheels, you get (2*(1-2*0.2)/(1-0.2))=150% as much torque at the wheels.

If you had two engines, one with 300lb-ft of torque at 5000rpm, and one with 150lb-ft at 10000rpm, geared to the same mph at the top of 1st gear, and assuming they had the same rotational inertia (which is likely given the same power output), the first one would accelerate much more quickly at low vehicle speeds.  This difference would get smaller at higher vehicle speed and lower (higher numerically) gear ratios, where the rate of change of engine speed is smaller.

Having said that, I 100% agree with your opening statement about confusing Tq vs HP.  At a given engine speed, torque is power (except for some unit conversions).  Low end torque equals low end power, high end torque equals high end power, and more average torque equals more average power.  The confusion lies in the fact that that peak torque always occurs at a lower rpm than peak power does.

I had not actually given though to the aspect of the rotating mass.  I'll try to run some calculations in the next couple of days on a couple of roughly standard setups... like a 14lb flywheel at 13" diameter and a 6lb flywheel at 13" diameter and some crank calculations as well to see what amount of power is actually required for those situations.  I'll also have to look at actual gearing etc to determine what rate of difference in the rpm acceleration alone is occuring.  Gear to gear there won't be a huge difference but, 4th gear to 1st gear there will be.  The one aspect that comes into play there though is that in a higher horsepower/torque car often the parastic load is of little consequence in the 1st gear and sometimes the 2nd as the vehicle's output well exceeds the capability of the tires and thus the losses from the faster windup trying to accelerate the rotating assembly become somewhat inconsequential.  I think one assumption that you made that I'm not sure I agree with is with is the rotational inertia.  Now I agree that at 5,000 rpm's for the first engine it may have roughly the same inertia as the 10,0000 rpm engine at those speeds, or we can at least assume that, especially if you use the basic formula of P(momentum)= mv wherein if P is equal to one another the velocities are defined you can see that the higher velocity engine (velocity is not really right, this should be a rotational issue, but this simplifies it for discussion) has a much smaller mass to be equivalent.  Therefore, the amount of of torque that would need to be applied to the rotating assembly of the 10,000 rpm motor would likely be LESS to accelerate it up another 100 rpm's in a 1 second than the 5,000 rpm engine.  Reason being that given that they have the same inertia the 10,000 rpm motor having less mass, and then simplifying the acceleration factor based upon  F=ma it will take less force to accelerate the 10,000 rpm engine, and therefore less torque would be necessary to do so than for the other motor in a similar situation, meaning that they could theoretically both accelerate nearly similarly with regards to the rotational mass of the engine.  By the nature of the mechanical forces involved it is typically a safe assumption to say that a high revving motor has less rotating mass, hence the reason that when people work to create a high revving engine they typically must lighten every aspect of the rotating assembly along with adjustments to piston size and stroke to further reduce weight and piston ring speed relative to the wall etc.  Regardless though, you bring up a great point and you now have me curious in a more practical example like a street car to what extent the loading of the rotating mass has an effect.

 Later!

 Steve

http://www.mazda6tech.com/files/wheel_inertia.xls

 This spreadsheet addresses inertia for wheels/tires and flywheels but could probably be massaged to do more.

djsilver:

http://www.mazda6tech.com/files/wheel_inertia.xls

 This spreadsheet addresses inertia for wheels/tires and flywheels but could probably be massaged to do more.

I didn't go into the wheels section and check but the math is wrong on the flywheel spreadsheet.  The "R" value is supposed to be a radius, they used the diameter in the spreadsheet. It causes the values to be off by a factor of 4 which has obviously a HUGE impact in what the results are.

Back to the original topic - I think the best metric would be area under HP curve over a given rpm-ratio. It is not enough to specify area under HP curve vs rpm. For example area under hp curve of 3000-5000rpm vs 7000-9000 are both over a 2000 rpm interval, but the ratios are 1.67 vs 1.28. Essentially the lower rpm range covers more than "twice" the effective range of the 7-9k range. The ratio of importance can be determined by either looking at the gearbox and taking the ratio of 1st to 2nd gear, or alternatively if you desire to avoid downshifting the ratio of redline speed in 2nd to minimum speed at which point you will downshift. For a CVT its clear this ratio is effectively 1.0 and you are only concerned with peak power. Typical ratios would seem to be in the 1.4-1.7 range.

 This is a simple numerical integral of the hp curve - and would not be difficult to add to the computer readout of a typical dyno. The number would be a great way to "yardstick" an engine relative to another.

underpaidslave:

Back to the original topic - I think the best metric would be area under HP curve over a given rpm-ratio. It is not enough to specify area under HP curve vs rpm. For example area under hp curve of 3000-5000rpm vs 7000-9000 are both over a 2000 rpm interval, but the ratios are 1.67 vs 1.28. Essentially the lower rpm range covers more than "twice" the effective range of the 7-9k range. The ratio of importance can be determined by either looking at the gearbox and taking the ratio of 1st to 2nd gear, or alternatively if you desire to avoid downshifting the ratio of redline speed in 2nd to minimum speed at which point you will downshift. For a CVT its clear this ratio is effectively 1.0 and you are only concerned with peak power. Typical ratios would seem to be in the 1.4-1.7 range. This is a simple numerical integral of the hp curve - and would not be difficult to add to the computer readout of a typical dyno. The number would be a great way to "yardstick" an engine relative to another.

I'd suggest using the torque curve instead as that directly correlelates to thrust. 

Torque is nearly meaningless without a lot more information than the metric I proposed above. 500ft-lbs of torque may seem like a lot, but if it is from an electric motor at 0.1 rev/sec , its worthless (except perhaps as a winch). While the statement torque directly relates to thrust is true, the same statement is true about horsepower - only its far simpler. Give me just HP and velocity I can give you thrust with one immediate calculation. If you give me torque  I need to know the gear ratio, the tire-diameter, and the final drive in order to give you a thrust figure.


The two methods are mathematically equivalent - however the original poster wanted to characterize cars based upon a single number that is more useful than peak HP. HP is just far easier to deal with than torque in these calculations.

Stan:

underpaidslave:

Back to the original topic - I think the best metric would be area under HP curve over a given rpm-ratio. It is not enough to specify area under HP curve vs rpm. For example area under hp curve of 3000-5000rpm vs 7000-9000 are both over a 2000 rpm interval, but the ratios are 1.67 vs 1.28. Essentially the lower rpm range covers more than "twice" the effective range of the 7-9k range. The ratio of importance can be determined by either looking at the gearbox and taking the ratio of 1st to 2nd gear, or alternatively if you desire to avoid downshifting the ratio of redline speed in 2nd to minimum speed at which point you will downshift. For a CVT its clear this ratio is effectively 1.0 and you are only concerned with peak power. Typical ratios would seem to be in the 1.4-1.7 range. This is a simple numerical integral of the hp curve - and would not be difficult to add to the computer readout of a typical dyno. The number would be a great way to "yardstick" an engine relative to another.

I'd suggest using the torque curve instead as that directly correlelates to thrust. 

 

 That's part of what we are getting at though in the discussions... you DON'T want to use the torque curve.  The gearing will actually be the measure of your thrust by being a multiplier and converter of the horsepower into the actual motive force for the car.  Therefore, it is the area under the horsepower curve that actually matter. 

 I am struggling though to get my head around the previous post as to why the lower rpm range is giving twice the "effective range.  That is what I'm a little unclear on.  Any math to elaborate on that underpaidslave would be great as I'm not making the connection right now.

underpaidslave:

Torque is nearly meaningless without a lot more information than the metric I proposed above. 500ft-lbs of torque may seem like a lot, but if it is from an electric motor at 0.1 rev/sec , its worthless (except perhaps as a winch). While the statement torque directly relates to thrust is true, the same statement is true about horsepower - only its far simpler. Give me just HP and velocity I can give you thrust with one immediate calculation. If you give me torque  I need to know the gear ratio, the tire-diameter, and the final drive in order to give you a thrust figure.


The two methods are mathematically equivalent - however the original poster wanted to characterize cars based upon a single number that is more useful than peak HP. HP is just far easier to deal with than torque in these calculations.

 Part of the point that I have been making is that the presence of the gear box is what greatly confuses the situation and why torque falls from value in discussion for the most part.  It can get complicated quick, but because gearing can be used to multiply torque up for use, there is essentially no mechanical adjustment that can increase your horsepower.  Therefore, it is the overall "quantity" of horsepower available to be worked with that then results in which vehicle/engine would typically be faster as long as the assumption that the gearbox is optimized to the application and the engine is in place. 

TurfBurn:

Stan:

underpaidslave:

Back to the original topic - I think the best metric would be area under HP curve over a given rpm-ratio. It is not enough to specify area under HP curve vs rpm. For example area under hp curve of 3000-5000rpm vs 7000-9000 are both over a 2000 rpm interval, but the ratios are 1.67 vs 1.28. Essentially the lower rpm range covers more than "twice" the effective range of the 7-9k range. The ratio of importance can be determined by either looking at the gearbox and taking the ratio of 1st to 2nd gear, or alternatively if you desire to avoid downshifting the ratio of redline speed in 2nd to minimum speed at which point you will downshift. For a CVT its clear this ratio is effectively 1.0 and you are only concerned with peak power. Typical ratios would seem to be in the 1.4-1.7 range. This is a simple numerical integral of the hp curve - and would not be difficult to add to the computer readout of a typical dyno. The number would be a great way to "yardstick" an engine relative to another.

I'd suggest using the torque curve instead as that directly correlelates to thrust. 

 

 That's part of what we are getting at though in the discussions... you DON'T want to use the torque curve.  The gearing will actually be the measure of your thrust by being a multiplier and converter of the horsepower into the actual motive force for the car.  Therefore, it is the area under the horsepower curve that actually matter. 

 I am struggling though to get my head around the previous post as to why the lower rpm range is giving twice the "effective range.  That is what I'm a little unclear on.  Any math to elaborate on that underpaidslave would be great as I'm not making the connection right now.

You busted me but it is simple. Say in engine #1 60mph was at 5000rpm for motor #1 and 9000rpm was for motor #2. The speed for 3000rpm for motor #1 = (3000/5000)*60 = 36mph, the speed for 7000rpm for motor#2 is (7000/9000)*60 = 46.7mph

That 2000rpm range "buys you" 24mph in motor #1, but only 13.3 mph for motor #2 - so I lied and its not more than twice the "range" its 24/13.3 = 1.8x the range (just less than twice). I made a hasty head math error above.

The ratio of speeds in each gear is of course set by the transmission, why do you think the F1 cars have wildly close gear ratios in top 5-6 gears? Because its faster to have a narrow powerband and make use of it via gears.

Hope this helps

I suspected it was something along those lines, just wasn't sure how the math was coming in.  Basically given two fixed ranges obviously the gearing can potentially shorten the time that it actually can be applied.  The motor running 7k to 9k though will obviously have much less engine torque than the 5-7K engine if the horsepowers are matched.  This gets back in to one of the points I made earlier that due to the change in gearing it is sometimes actually good to run the gearing well outside of the "effective" areas of the power curve because of the greater advantage provided by the multiplier in the lower gears.

TurfBurn:

Stan:

underpaidslave:

Back to the original topic - I think the best metric would be area under HP curve over a given rpm-ratio. It is not enough to specify area under HP curve vs rpm. For example area under hp curve of 3000-5000rpm vs 7000-9000 are both over a 2000 rpm interval, but the ratios are 1.67 vs 1.28. Essentially the lower rpm range covers more than "twice" the effective range of the 7-9k range. The ratio of importance can be determined by either looking at the gearbox and taking the ratio of 1st to 2nd gear, or alternatively if you desire to avoid downshifting the ratio of redline speed in 2nd to minimum speed at which point you will downshift. For a CVT its clear this ratio is effectively 1.0 and you are only concerned with peak power. Typical ratios would seem to be in the 1.4-1.7 range. This is a simple numerical integral of the hp curve - and would not be difficult to add to the computer readout of a typical dyno. The number would be a great way to "yardstick" an engine relative to another.

I'd suggest using the torque curve instead as that directly correlelates to thrust.

 

 That's part of what we are getting at though in the discussions... you DON'T want to use the torque curve.  The gearing will actually be the measure of your thrust by being a multiplier and converter of the horsepower into the actual motive force for the car.  Therefore, it is the area under the horsepower curve that actually matter.  I am struggling though to get my head around the previous post as to why the lower rpm range is giving twice the "effective range.  That is what I'm a little unclear on.  Any math to elaborate on that underpaidslave would be great as I'm not making the connection right now.

I'd still use the torque curve for normal cars.   As I wrote I thought we were talking about THRUST.   Which also requires we deal with gears, tire size, traction, weight/mass and so forth...inherently...in typical cars which change RPM as they gain speed in a given fixed ratio gear.  

I datalog all the time.  When you plot longitudinal Gs versus MPH for example you see the strength of the thrust (in Gs) versus speed.  Accounting for things like cold air intakes, air resistance, rolling resistance and whatever else as this is the net result.  The MPH can be converted to RPM.  The Gs can be converted to HP with some more information.  

You also notice that the SHAPE of this plot is...the same as the torque curve.  It seems like it might be easier to understand for more people.  

Here is such a plot for a car accelerating in 2nd gear 20-45 MPH with various intakes.   This speed range is common at some local event on smaller sites.  The shapes of the curves corresponds to dyno torque curves.  I prefer the datalog tests as they account for more variables though.  The red trace is with still another intake and 3rd gear.  Of course the plot is lower due to gearing. 

Take a look at the green curve and the black curve.  The green has more peak thrust than the black.  But in the lower mph ranges the black has more area under the curve.  In the higher ranges it's all green.  At the site I am thinking about...where you only briefly go over 40 MPH...the green setup might tend to be best as it will pull harder out of all the 25-35 mph course elements.  

Of course...the blue setup is best of all for this car...strongest low and high end with a good middle.  

Stan:

I'd still use the torque curve for normal cars. As I wrote I thought we were talking about THRUST. Which also requires we deal with gears, tire size, traction, weight/mass and so forth...inherently...in typical cars which change RPM as they gain speed in a given fixed ratio gear.

I datalog all the time. When you plot longitudinal Gs versus MPH for example you see the strength of the thrust (in Gs) versus speed. Accounting for things like cold air intakes, air resistance, rolling resistance and whatever else as this is the net result. The MPH can be converted to RPM. The Gs can be converted to HP with some more information.

You also notice that the SHAPE of this plot is...the same as the torque curve. It seems like it might be easier to understand for more people.

Here is such a plot for a car accelerating in 2nd gear 20-45 MPH with various intakes. This speed range is common at some local event on smaller sites. The shapes of the curves corresponds to dyno torque curves. I prefer the datalog tests as they account for more variables though. The red trace is with still another intake and 3rd gear. Of course the plot is lower due to gearing.

Take a look at the green curve and the black curve. The green has more peak thrust than the black. But in the lower mph ranges the black has more area under the curve. In the higher ranges it's all green. At the site I am thinking about...where you only briefly go over 40 MPH...the green setup might tend to be best as it will pull harder out of all the 25-35 mph course elements.

Of course...the blue setup is best of all for this car...strongest low and high end with a good middle.

Couple of issues here : 

I posted in response to the OP's dissatisfaction with the "standard" way of comparing engines via a pair of numbers (peak hp / peak tq). I proposed the alternate metric of area under HP curve over a ratio of RPM (relevant to the application)  (as the above post indicates over a fixed rpm range isn't sufficient). This "distills" the important characteristic of an engine into one number (obviously one number is less exact than a table ie. curve). Any attempt to use to torque only to do the same thing would require a LOT more info (gear ratios tire diameter etc etc) to achieve the same descriptive power. The HP # is more quickly usable as it is independent of the gearing.

 Nice plots , I want the blue curve and if not blue then gold! :)

When you talk curves of hp or torque - there is no difference in descriptive power choosing one over the other. Most drum dynos actually measure HP and calculate torque - it isn't important the 2 curves are 1-1 related mathematically, given one the other is easy to get.

The only thing I disagree with you is  :

Stan:

As I wrote I thought we were talking about THRUST. Which also requires we deal with gears, tire size, traction, weight/mass and so forth...inherently...in typical cars which change RPM as they gain speed in a given fixed ratio gear.

 Using HP allows us to NOT have to deal with tire size/gear , but merely require the mass and the speed.

For example in your above plots - if the "blue" car in question weighed 2200lbs (as tested - just making math ez for my head) with an 0.34g acceleration at 35mph would indicate roughly 70 net hp. No need to worry about gearing at all! At 45mph the blue car is making 94hp net. By net I mean minus the drag.

With some "coast down" measurements of deceleration vs speed you could build a drag model and correct for that.

 given s = vehicle speed in mp/h
         a = acceleration in g's (at speed s)
         w = vehicle weight in lbs

 HP = (1/365) * s*a*w,  If you know any 3 of weight, speed, acceleration, hp - you can easily solve for the fourth without "knowing" the gearing or tire size.