Stupid and unrealistic "Which is faster?" question

In reply to SVreX:

I'm trying to figure out which motor would be better/come up with a good reason to build a destroked motor.

In reply to Swank Force One:

Only a few reasons to destroke.

1. Fun, if you find a car that drives like an old 2 stroke dirt bike fun. My Yugo is kinda like that, something like a 80mm bore and a 55mm stroke. Consequently it revs like a chainsaw.

2. Higher revs to get more power if rules fix displacement.

3. Crank becomes harder to break(less moment arm in the crank throw). This just trades off for higher tensile stress on the rods though.

Swank Force One wrote: In reply to SVreX: I'm trying to figure out which motor would be better/come up with a good reason to build a destroked motor.

Wouldn't this be like comparing a modified 4.8 to 5.3 LS engines?

Destroking a motor isn't going to raise your rev limit enough to compensate for the smaller displacement. If you have 3 LS motors- a 4.8, a 5.3 and a 6.0, and spend the same care and attention on each, the 4.8 will be slower than the 5.3, which will be slower than the 6.0.

Period. I will accept no argument.

In reply to Swank Force One:

What are the transmission gear ratios in your MX6?

Streetwiseguy wrote: Destroking a motor isn't going to raise your rev limit enough to compensate for the smaller displacement. If you have 3 LS motors- a 4.8, a 5.3 and a 6.0, and spend the same care and attention on each, the 4.8 will be slower than the 5.3, which will be slower than the 6.0. Period. I will accept no argument.

True, but if you can't have 6 liters, only 4.8, you can make more power by making it rev faster, making it pump more air(thus more fuel burned), and the way to make that work without grenading it is to reduce stroke. The extreme example would be F1 where the engines can't rev any higher than they do because the fuel won't burn as fast as the piston moves.

Swank Force One wrote: And i'm now confused about all this, because it would seem that the math isn't explaining my question. Looking at this dyno chart, if the "torque is everything" sect is correct, then this car would be fastest through the quarter mile short shifting at around 4500-5000rpms. (MX6, and yes, i realize that this is not a 600hp motor, nor are we comparing it to one.) It's not. It's fastest shifting at redline, which doesn't even drop down far enough to ever hit peak torque. Why is this?

I will almost certainly regret posting in this quagmire of (both good and bad) information...But that's never stopped me before.

A lot of people have tried to over complicate a simple question. The answer comes down to nothing more than a matter of which has more torque being applied to the ground, and all goes back to bradyzq's reply on page 1. Seriously, there's nothing more to it than that.

Note that toque on a chassis dyno may be measuring at the wheels, but what it's actually displaying is "engine torque minus drivetrain efficiency losses" regardless of people calling it “at the wheels”. The all important torque at the ground, is the dyno number multiplied by the transmission ratio, multiplied by the final drive ratio, and divided by tire rolling radius (as measured in feet).

The only measurable characteristics that define this particular discussion are torque, rpm, and gearing. While horsepower is a theoretical concept that makes it easier to discuss engine characteristics in general, and is important for understanding that equal hp is going to be equal, it’s not particularly useful to the question at hand since it’s an unmeasurable (for all intents and purposes ‘imaginary’) number, merely being derived from torque and rpm.

A quick search brought up the Mazda H5M-R w/ 2.2T gear ratios as: 3.250, 1.772, 1.194, 0.926, 0.711.

Since the tires and final drive used are the same regardless of which gear you’re in, we can drop them from the calculations entirely and focus only on the transmission ratios. So based on your transmission ratios and the shown 6000rpm dyno redline, the transmission output torque in each gear would then become: 731, 399, 269, and 208 in 1st through 4th gears. Fifth does not matter, since there is no shift point after that.

After each shift, you would end up at roughly the following rpm: 3270, 4040, 4655, and 4605 in 2nd through 5th gears. Doing the same analysis as above at the post-shift rpm, the transmission output torque would be approximately: 532, 519, 352, 274 after the shift from redline. This means you would only be better off shifting right as you hit redline on the 1-2 shift. After that, you’ll be putting more torque through the tires at the same wheel speed by shifting earlier. Doing it visually off your chart is a bit of trial and error, but the physics appear favor shifting your car at around: 5400 for the 2-3 shift (3640rpm into 3rd), 5000 for the 3-4 shift (3880 rpm into 4th), and 5000 again on the 4-5 shift (3840 rpm into 5th).

This is simply looking at torque on your graph at a given rpm, and multiplied by the transmission gear ratio. As a few people have repeatedly stated, the same basic math applies to your original question, once you convert hp back into torque at that rpm.

With enough rpm and a short enough gear ratio, you can make the fastest and quickest car in the world with 1 ft-lb of torque.

With enough torque and a tall enough gear ratio, you can make the fastest and quickest car in the world with 1 rpm.

Don’t get caught up on one necessarily being ‘better’ than the other. Either method is capable of being faster than the other and each is only advantageous if used correctly.

In reply to Driven5:

Not sure. Not great in my opinion. But that's not the car i'm talking about.

Kenny_McCormic wrote: In reply to Swank Force One: Only a few reasons to destroke. 1. Fun, if you find a car that drives like an old 2 stroke dirt bike fun. My Yugo is kinda like that, something like a 80mm bore and a 55mm stroke. Consequently it revs like a chainsaw. 2. Higher revs to get more power if rules fix displacement. 3. Crank becomes harder to break(less moment arm in the crank throw). This just trades off for higher tensile stress on the rods though.

Well... there's also the theory in this case that the head would pair better with a smaller motor.

In reply to Swank Force One:

I know it's not the original car in question, but understanding the answer to the issue you brought up with it applies similarly to understanding the answer to the original question.

Ah ok. I'm digesting your long post. I appreciate it, will respond tomorrow after I sleep on it.

As a follow on, the way I came to those shift points, was looking for the rpm that creates equal transmission output torque (after gear multiplication) in both the current gear AND the next higher gear after the rpm drop. It's not exactly intuitive, but hopefully will make sense. Thus at a lower rpm than this point the tires would still be getting more total torque in the current gear, but continuing to a higher rpm means that the tires would actually be getting more total torque in the next higher gear.

In reply to Driven5:

HP is not a theoretical or imaginary number. It is real. It is power. It is usually expressed in watts (ya know, like a lightbulb). Just because there is an equation and some other items factored in, doesn't make it imaginary.

Story of Hp. A single HP comes from a horse lifting 440 lbs 100 feet per minute. There are 746 watts in 1 hp.

Horsepower IS torque over a time period.

Again:

Torque tells you how heavy of a load you can pick up.

Horsepower tells you how quickly you can pick that load up.

Even in this little example, it is obvious that horsepower is what you are interested in.

wvumtnbkr has it totally correct. After reading through this thread, I think it might help anyone still not totally grasping the concept to remember this too: given limitless gearing options (CVT) power is all that matters, it will accurately predict acceleration no matter what the weight being discussed.

However, as gearing becomes more limited (fewer gears), and weight gets greater, torque will become more important to overall vehicle performance, because gearing multiplies engine torque at a given engine speed.

So it is not just the area under the curve (known mathematically as the integral) of the engine in question; it is the area under the curve of the available power of the engine, along with the available multiplication of torque by all of the transmission's gears and differential.

To maximize acceleration you would calculate available friction, then work from there to keep torque at the contact patch just below that figure using the combinations of gearing and power available to you.

Sorry for the long post. J

My apologies to all for using the word "theoretical" when "purely mathematical" would have been more accurate.

Don't miss the forest for the trees.

Driven5 wrote: As a follow on, the way I came to those shift points, was looking for the rpm that creates equal transmission output torque (after gear multiplication) in both the current gear AND the next higher gear after the rpm drop. It's not exactly intuitive, but hopefully will make sense. Thus at a lower rpm than this point the tires would still be getting more total torque in the current gear, but continuing to a higher rpm means that the tires would actually be getting more total torque in the next higher gear.

I see what you've done here, but my experience has been that shifting for greater torque in next gear isn't the quickest. Shooting for more HP in the next gear does this well, or maximizing the area under the HP curve, which I have seen work better.

This calculator is the best one I have seen for it online. I use a similar method with more points, but it tells the story decently.

http://www.bgsoflex.com/shifter.html

The only tweak I have to make it to adjust for extra rpm drop due to shift speed and on drag cars on the 3-4 shift to shift early or late depending on rpm in the traps. Shifting late if you can do it fast, usually works out better.

So the shortest gear (most torque) will always provide the best acceleration?

Just stay in first gear and you will win the drag race!

Again, the variable missing is time. Horsepower IS torque (with respect to time).

Torque is instantaneous. Horsepower is the same thing spread over time.

Paul_VR6 wrote: This calculator is the best one I have seen for it online. I use a similar method with more points, but it tells the story decently. http://www.bgsoflex.com/shifter.html

Interesting that it outputs both the 'same horsepower' (same total torque) shift points and 'area under the curve' shift points. The 'same horsepower' calculation is the one I just did above and gives the same answers I came to, regardless of whether setting it up to use horsepower or torque as the inputs. At the moment after the 'same horsepower' shift point, this would certainly be the better accelerating option. I'll concede though that the 'area under the curve' method may still accelerate more quickly overall, if it provides more torque to the ground more of the time through the entirety of each gear.

The point however, is that torque applied to the ground is only thing propelling the car forward at any moment in time regardless of speed or rpm, and continuously maximizing it produces the greatest acceleration. Period.

What is your math on that? I am getting a much different result. On the last motor I set up I am making all the shifts well past torque and hp peak and the torque is nearly doubling after the shift...all at the same hp before and after.

Even looking at your example above, depending on the powerband of the engine, those shift points look very early (stump puller?)

I would agree that the MX6 dyno posted by SFO could be defined as a bit of a stump puller. In fact, it makes for a great example to use since the 'same horsepower' and 'area under the curve' methods from that calculator produce such wildly differing results. Most engines used in a way that warrants worrying about shift points, make enough torque up top that either method will simply tell you to shift at or near redline in most gears.

Having had a chance to think it over for a while, and roll some numbers around in excel, I've reconfirmed my belief in the 'same horsepower' (same applied torque) shift point methodology that I used above producing greater acceleration. The 'area under the curve' methodology focuses on the rev range within each gear, but doesn't appear to account for the increase in area able to be utilized by going into the next gear.

Honestly I'm pretty surprised there aren't more people on here who can explain this well.

I've ranted way too long about this before so I'll try to keep this short, otherwise google it.

Firstly, power is all that matters. Period. You can stop reading now. But by definition Power=Work/time=ForceDistance/time=F/velocity=masaccelerationvelocity. Alternatively expressed for a rotational system: Power=Work/time=Torque# Rotations/time=Torqueomega. Right, so P=mav and if we have two cars of equal mass moving at the same speed. The only way for one to out accelerate the other is if it has more power. Since P=Tw you can always trade torque for speed or the reverse thru gearing but fundamentally you can't exceed the amount of power the engine has. Power is all that matters.Related to your original question, this is why a car making X Torque at Y RPM and a car making Y Torque at X RPM are just as fast. They make equal peak power and if you dictate that they are driving the same speed(at peak/any equal power) then they MUST make the same torque at the rear wheels. If you take your hypothetical step further and imagine two cars with engines that hold a constant power over the entire rpm range. No matter what gear or rpm the cars are at, if they're moving the same speed then they will accelerate at the same rate, this is because they are still making the same power, regardless of rpm. P=av

Theres a good post somewhere on the internet, I'll try to find it.

The reason that the power vs torque debate has never been laid to rest, and never will be, is that when used correctly they are both equally valid and accurate methods of representation...Regardless of what each 'side' wants to believe. This is due to power being directly derived from torque. If you follow the torque calculations forward at any given point in time, you'll necessarily end up with power. And if you back calculate from power at the same point in time, you'll necessarily end up with torque. Basically power tells you what is happening, torque tells you why it's happening. Personally, I believe the less abstract concept to be the more effective descriptor. But as repeatedly illustrated since page 1 of this thread, obviously some people's minds think more like mine and others don't. In reality though, among those who genuinely understand how both work, it's just six in one and half a dozen in the other.

I think the discussion of shift points is funny...

Graph acceleration in all forward gears, you can measure this, or do it mathematically. Where the accel plots cross is where you shift.

It's hard to graph acceleration of a completely made up scenario.

I don't disagree that you shift right around max accel crossover between two gears, but you need to account for the shift time. Longer time to shift, the later you shift, generally.

This is like first grade and trying to read a book without pictures.