Graphs: torque in gears / rpm per mph in gears. 2.5L 6-speed auto trans.

I did not say anything like that. You fail to understand the concept.

What I said, in effect, was that the car would accelerate harder if it did not have an extra bog/pause at each upshift because there was a large step down in thrust (wheel torque) in each higher gear.

The graph is somewhat misleading.
The gap in real wheel torque between gears is not nearly as large (could even be non-existent) as the graph makes it seem.

When you're quickly accelerating in a low gear a lot of the engine power goes to spinning the engine and drive train up to speed. This means that the wheel torque applied to the ground (especially in low gears) when accelerating is a lot lower than what the graph shows.
 
The graph is somewhat misleading.
The gap in real wheel torque between gears is not nearly as large (could even be non-existent) as the graph makes it seem.

When you're quickly accelerating in a low gear a lot of the engine power goes to spinning the engine and drive train up to speed. This means that the wheel torque applied to the ground (especially in low gears) when accelerating is a lot lower than what the graph shows.

You make a great point. However, try a thought experiment. Jack the car up on jackstands. Put the car in 1st gear manually, and floor it. It'll take much less than 1 second to hit redline. And that's accelerating the driveline more than 8 times faster than in a 0-60 pass. At the actual rate of RPM increase, the power needed to accelerate the drivetrain is exponentially less (well, square/cube or somesuch). Anyone have links to info on this? I'm guessing drag racers might have info on heavier/larger radius tires vs smaller. I'll go look...
 
From: http://hpwizard.com/rotational-inertia.html#answers

"Just give me the answers

Shaving a pound from your tires is equivalent to shaving at most 2 pounds of non-rotating weight. That's PER TIRE, so a pound off each tire could worth close to 8 pounds of weight reduction. For wheels, the multiplier is closer to 1.6, so saving 5 pounds per wheel (20 total) would feel like a static weight reduction of 32 pounds. For brake discs, it can be as low as 1.2. Regardless of the equivalent weight ratio, you're best off reducing weight as much as possible, as you might expect. For flywheels...you'll have to read the detailed section. Sorry."

This seems like too little effect to me... :)
 
Here's a page where you can calculate the HP absorbed by a flywheel: https://www.w8ji.com/rotating_mass_acceleration.htm

"The wheels also speed up and slow down gradually. With an 11-second car, we have 11-seconds to speed the wheel up. Most of the horsepower pushed into the wheel and stored is pushed in near the end, when acceleration is least. Since we have more time to push the bigger amount of energy into the wheel, it takes less horsepower than we might expect."

For a flywheel, going from zero RPM, to 6,000 RPM (which is orders of magnatude greater than a tire): "If we have two seconds to spin up to 6000, the engine will push either 6 horsepower average with aluminum, or 12 horsepower average with steel, into the flywheel. If we have 6 seconds to spool up, average horsepower is either 2 or 4 horsepower."

I'm kind of bummed, now, because my 13 pound lighter wheel/winter tire package arrives tomorrow, and I won't be able to feel an acceleration difference!
 
From: http://hpwizard.com/rotational-inertia.html#answers

"Just give me the answers

Shaving a pound from your tires is equivalent to shaving at most 2 pounds of non-rotating weight. That's PER TIRE, so a pound off each tire could worth close to 8 pounds of weight reduction. For wheels, the multiplier is closer to 1.6, so saving 5 pounds per wheel (20 total) would feel like a static weight reduction of 32 pounds. For brake discs, it can be as low as 1.2. Regardless of the equivalent weight ratio, you're best off reducing weight as much as possible, as you might expect. For flywheels...you'll have to read the detailed section. Sorry."

This seems like too little effect to me... :)

that site has a little simplified calculator at the bottom of the page.
Probably not the most accurate thing, but I'm sure it gives reasonable numbers.

According to that little calculator, in first gear, the "equivalent mass" of a CX-5 (AWD gear ratios) is 1.72 times higher than the real mass of the car.
In second gear, it's just 1.26 times higher.

This difference in "equivalent mass" is probably enough to make the lines in your graph cross under acceleration.
 
But as posted earlier, most of that equivalent mass is the clutch, flywheel, and other *engine* rotating mass, which is always a factor. (Because those parts accelerate to 6,200 RPM, whereas the wheels/tires only rotate 720 revolutions PER MILE or 90 revolutions in 1/8th mile, which would have the car over 60 mph.)

Does anyone have speed/time logging out there? If my car were broken-in, I could video the speedo, and log each frame's speed, to see how closely the speed increase matched the calculated torque in gears...
 
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But as posted earlier, most of that equivalent mass is the clutch, flywheel, and other *engine* rotating mass, which is always a factor. (Because those parts accelerate to 6,200 RPM, whereas the wheels/tires only rotate 720 revolutions PER MILE or 90 revolutions in 1/8th mile, which would have the car over 60 mph.)

Does anyone have speed/time logging out there? If my car were broken-in, I could video the speedo, and log each frame's speed, to see how closely the speed increase matched the calculated torque in gears...

right, but in 1st gear, the engine spins up to 6200RPM very quickly. It does so much more slowly in 2nd gear, and even more slowly in 6th gear.
the actual wheel torque under acceleration will look something like this:
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Your torque/mph in gears graph is correct, but it leads some ppl (ColtX-5 for example) to conclude that closer gearing would result in faster acceleration.
 

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There's something wrong with that dyno chart... It seems to show that torque (equal to the car's acceleration, equal to the push-in-your-backside) is higher in higher gears. That's clearly not the case. I wonder if the clutch was slipping in the lower gears, or if the calculations that the dyno PC uses isn't taking into account the higher drum speed.
 
There's something wrong with that dyno chart... It seems to show that torque (equal to the car's acceleration, equal to the push-in-your-backside) is higher in higher gears. That's clearly not the case. I wonder if the clutch was slipping in the lower gears, or if the calculations that the dyno PC uses isn't taking into account the higher drum speed.

I don't think there is anything wrong with the chart.
The dyno PC likely takes info account the higher drum acceleration and compensates for it, but it doesn't correctly compensate for the inertia of the engine/transmission/wheels.

If you fit a lighter flywheel, or even different rear-end ratio to a car you'll end up with different dyno numbers, but in reality the engine is still producing identical torque.

Here is a quote I found on another forum:
Here's the reason, I'm quoting Patrick G on LS1Tech.com:
"Higher ratio gears will give you lower chassis dyno numbers for a strange, but logical reason. In essence, the DynoJet calculates hp based on the time it takes to spin up the 2800lb roller assembly. It's basically work divided by time and rpm. Think about this: If you car is at idle in neutral and you stab the throttle, it will take time to accelerate to redline...let's say 1.1 seconds. Now let's say it takes 8.2 seconds for your car to accelerate the DynoJet from low speed to top speed with 3.23 gears and 7.3 seconds with 3.73 gears. Dyno printout says 355 rwhp with 3.23 gears and 346 rwhp with 3.73 gears...why?

Think aabout this: In the 8.2 seconds it takes to spin the rollers with 3.23 gears, it would still take the motor about 1.1 seconds to overcome its own inertia (idle to redline). There's about 13.4% of the work used just to accelerate the motor itself. With 3.73 gears, the time to reach redline decreases to 7.3 seconds. Divide the 1.1 seconds into the 7.3 seconds and you will see that overcoming the internal engine inertia costs 15.1% of the work with 3.73 gears. There is less hp available during this time period to spin the rollers so the DynoJet will read a slightly lower hp figure. Make sense, or did I lose you?"
 
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That makes sense, and I think I was looking at the chart from the wrong 'direction'. This has been a very informative thread, thanks to all who participated!
 
That makes sense, and I think I was looking at the chart from the wrong 'direction'. This has been a very informative thread, thanks to all who participated!

I'm probably beating a dead horse at this point..
(deadhorse)

I've thought about this some more and closer ratios would definitely help with climbing hills, with a full car, and especially while towing uphill.
The inertia of the engine in those situations doesn't matter as much (or at all)

The only tine they wouldn't help is with WOT acceleration on flat ground with an empty car. (what magazine tests do)
 
If you look back at the torque curves, and interpolate between the data points of 2nd and 3rd gear, and imagine a 2.5 gear, the torque amount would not appreciably increase. Maybe I can throw another gear in the graph, and post it tomorrow...
 
LOL! The gaps in the gears are too wide. Max. acceleration occurs when the shift is made at the intersection of the wheel torque curve of one gear and the next higher. The next higher gear has more available RPM *and* at least the same wheel torque..

the gears are pretty far spaced apart.. you feel it during full throttle acceleration. but the gearing is very well calibrated for half-throttle application, revving to 3-4k RPM.
 
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