Short shifting in gt sport doesn't make sense.

(Torque x Rotational Speed x C) = (10 x Torque x 1/10 Rotational Speed x C) = Power

If you can't make sense of that then there is nothing anyone can do to help you understand this.

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I'm glad I just do the driving.:lol:
 
What about coasting before braking?

like others have said, it would be better for tires and probably smoother, would take some practice. Thing is though, be careful in traffic because others may not expect you to do that so it could cause more incidents. it also could make you more susceptible to overtakes.
 
I have no idea what you are trying to illustrate, but, if you cannot see that you have changed 100Nm to 1000Nm (which is indeed changing the power) to somehow support your math, there is nothing anyone can do to help you understand this.

Nm is a measure of torque - not power.

Electric motors produce a nearly flat TQ curve from 1rpm on, and is why they do not need a gearbox. It's not that they produce low end or high end, rather virtually unchanged power continuous across it's operating range/rpm.

You’re mixing up torque with power again. The power curve of an electric motor is nearly flat, not the torque curve.

What? the example I quoted, was where you were adding multipliers to the TQ, then changing the gear ratio to solve some math issue.

No, the torque changed BECAUSE the gear ratio changed. The division of speed is equal to the multiplication of torque, which means that power is not changed. 200 bhp at the flywheel is 200 bhp at the powered wheel.

200 Nm at the flywheel is NOT 200 Nm at the powered wheel - it’s 200 Nm * gear ratio * final drive ratio.

All I was illustrating is that you cannot add a multiplier to the source power... it is simply TQxgear ratio, not "#"xTQ x gear ratio.
here, I will re-quote...

Look at the labels though. It says speed - not gear ratio. Torque * speed = power.

TQ at the flywheel is unchanged, the frictional losses will be measured at the wheels.

Nope, it’s connected by the clutch, so friction of the gearbox affects the torque at the flywheel as well. It’s one and the same system.

No gearbox/clutch on the planet will change the source power... it will effect the driven wheels.

What makes you think I’m manipulating the source power? I said that the gearbox is a torque multiplier and that power remains unchanged. Yet you somehow believe that I change the source power, whatever that means?
 
Nope, it’s connected by the clutch, so friction of the gearbox affects the torque at the flywheel as well. It’s one and the same system.

The actual torque output that an engine makes is measured at the crankshaft/ flywheel of the engine with none of the drivetrain attached and is advertised as such.

Usually the next measurements will be taken on a chassis dyno at the drive wheels and those numbers will be advertised as well.

All loss of hp/torque from the crankshaft figure to the rearwheel figure will then be classed as the frictional losses of the drivetain as a whole.

The clutch or the transmission will not affect or decrease the actual figures of the amount of power generated by the engine only that due to the drag of weight and frictional resistance of the clutch and transmission that less of those original power figures will still be available due to losses caused by the drivetrain.

So all loss from the drivetrain will still be an effect of the drivetrain eating up some of that output and not the engine actually generating less power at the same loads or rpms.
 
Okay guys please. I understand that you're trying to figure out the exact math to prove your... whatever it is. But the main concern for me is... where's the proof? Here is a link in a race with the amg gt3 (the key point here it's that it's real.) So look at the rev limiter and tell me if he is shifting half way to save fuel.
 
Okay guys please. I understand that you're trying to figure out the exact math to prove your... whatever it is. But the main concern for me is... where's the proof? Here is a link in a race with the amg gt3 (the key point here it's that it's real.) So look at the rev limiter and tell me if he is shifting half way to save fuel.

Hmm, try looking at LMP1s maybe? I'm not 100% sure, but they do have limitations on how much fuel they need to use over the course of a certain amount of laps. So there you'll see more fuel-saving. There's lifting, and maybe shortshifting
 
Hmm, try looking at LMP1s maybe? I'm not 100% sure, but they do have limitations on how much fuel they need to use over the course of a certain amount of laps. So there you'll see more fuel-saving. There's lifting, and maybe shortshifting
They do indeed short shift. When coming out of corners. I've watched plenty of onboards and live streams and have not been able to catch anyone short shifting for the entirety of the race.
 
I really want to understand this...
Why are you multiplying TQ by 10?
Why are you then dividing rpm by 10?
he is showing that the variables are inversely proportional to eachother.

All he is saying (mathematically) is that if you keep power constant, anything you do to torque will affect rpm
 
Below are just some parts of articles from different results from fairly recent IMSA series races that discuss using fuel and tire saving strategies from the real world races. I am guessing now that in real world road racing most fuel saving strategy is done via fuel map and computer rather than manually although I am sure both are used to some extent.

Hopefully this will show how fuel saving and fewer and shorter pit stops are a real world racing strategy and not just something done in game in the virtual races.


The Camaro that eventually won made a pit stop earlier and was able to pass Hindman on lap 43 with fresher tires, but Hindman was back in front for four more laps before he finally pitted on lap 68 for his one and only pit stop. In addition to saving fuel and his tires during his 1:07 stint on a track that was getting more slippery every lap, he also turned the car over to Cassels with no damage for the final push. The pit stop was fast, Cassels set very competitive lap times and successfully avoided any crashes too, and the team's reward was its fifth top-five finish of the season to date. When adding in the time during qualifying, Hindman's stint and the pit stop, the team stretched one set of Continental tires for approximately one hour and 20 minutes of competition
I just tried to save fuel, save tires, and tried to keep it in one piece for Cameron, but that was absolutely chaos.


In the #911 Porsche 911 RSR, however, Dirk Werner was on course for a podium spot. The fuel-saving strategy that his team had worked out for him after he took over the cockpit from Patrick Pilet eventually began to pay off. For long stretches he was running in third, and he even managed to advance to second place. At times he was less than a second off the leader, and in a thrilling final phase defended his position with a spirited duel against his strongest pursuers. However, shortly before the flag he had to let a Ferrari overtake him.

Neel Jani says LMP1 privateers are now faced with the “difficult” task of manually saving fuel per lap in order to not exceed the FIA’s energy limits for non-hybrid competitors.
“We had to do quite a lot of fuel saving per lap


In Protype challenge, the #38 car spent the least amount of time on pit road by attempted to make a 2 stop strategy work but the fuel saving required caused the #8 and the #52 to finish in front despite spending 40-60 seconds longer on pit road. In prototype the 1st and 3rd place finisher were able to do the race on only 2 stops. The #31 car also had a 2 stop strategy but has a 3rd stop that was the drive through penalty for contact with its teammate. The #70 Mazda spent almost a full minute longer on pit road than any of their competitors.
 
@VFOURMAX1 that makes a lot of sense. So you're basically saying that the "short shifts" in the game are already applied in real racing, but it is already implemented in the fuel map/ecu programming. However in gts you have to do it manually. I'm on my way to understanding short shifts in gts. Now all I need to do is look at the data and numbers. Can you link me to any sources by any chance?
 
However in gts you have to do it manually.

You can do it in GTS with fuel map changing during the course of a laps progression but it is just easier in game in my opinion to leave the fuel on map 1 full power and back off the go pedal where and as much as possible while still maintaining somewhat of a good pace as compared to those not employing the fuel saving strategy.

I am not sure of the technical rules for most of the car series but I do know up through at least last year and probably still current that Moto GP traction control was gps based and could be set independent for each individual corner.

Most racing these days the highest paid guy on the team is the guy in the pits with the laptop that analyzes the sensors data and programs the ecu and all the vehicles systems.

I do not know where to find a source for data and numbers. I would expect most race teams would keep their collected data pretty close to the chest and not share it for competitors to analyze.
 
Too many variables come into play to to use one rule of thumb as far as what RPM an engine will give the best performance in relation to dyno horsepower and torque numbers.

Displacement, bore, stroke all greatly affect where and how an engine makes power. Camshift lift and valve timing as far as whether the camshaft is advanced or retarded when installed will affect the power curve as well. Then we get into ignition timing, like I said so many different things control or can alter how an engine makes or delivers its power.

This is the reason that you can have two engines from the same manufacturer with the same cubic inch displacement and one will excel at under 5000 RPMS pulling the trailer to the track and the others strong point will be at 8500 RPM motivating a much lighter race car down the track.

Switch those two motors between the two vehicles and neither will perform adequately of the different task they would then be expected to perform.

Everything in the drivetrain and the vehicle it is installed in must be matched to all perform the same goal of the maximum performance of the task they were built to do.

Back when I use to drag race I preferred the big block torque and pull that I got out of the higher gears on the top end that I could not get out of the small blocks. The small blocks relied more on quick high revving rpms to make their power and if you tried to twist a big block that high you were just going to turn your engine into a grenade!

Again two different methods of trying to obtain the same goal which in this case was covering a 1/4 mile from a standing start in as short a time as possible.:cheers:

I completely agree, but in real life it does not change that fact that you must, absolutely must run the engine pretty far beyond RPM of max HP, not max torque, to optimize acceleration.
 
I really want to understand this...
Why are you multiplying TQ by 10?
Why are you then dividing rpm by 10?
A 10:1 combined gear ratio will give you 10 times the wheel torque while reducing the speed of the driven wheels by a factor of 10.
You can do more work, but slower. The rate at which work is done remains constant.
 
A 10:1 combined gear ratio will give you 10 times the wheel torque while reducing the speed of the driven wheels by a factor of 10.
First off, thank you for the straight response.
Absolutely 100% agree with your statement... which is how this all started out.
Remember...
1. That means that the power at the wheel is the same as the power at the flywheel.
2. 2 torque x 4 speed is the same power (8) as 4 torque x 2 speed.
I'll focus just these 2 statements, ignoring there were a multitude of misconstrued facts along the way, which have been conveniently edited...
1. You yourself have noted that power at the wheel is manipulated by gearing... first quote of this post... and is what i have said from the beginning.
2. The second example clearly applies a multiplier to TQ, then adjusts gearing to justify simple arithmetic... but has no relevance in a engine/transmission/diff/wheel power delivery system.
Before you "math" that, consider neither TQ nor HP are delivered in a linear form where rpm is directly proportional to output, rather "power" (used as a loose term, I'm sure eran will jump on that) is delivered in a logarithmic curve...

You can do more work, but slower. The rate at which work is done remains constant.
I'm rather certain this is incorrect... I mean simple math will defend this (x10 then /10), but...
Work is not constant when applied to a internal combustion engine installed in a vehicle with multiple transmission ratios...
Due to the "power" delivery curve, and the gear changes events, the operating ranges/windows differ relative to average "power" utilized over a specific range of rpm.
Work is only constant if the rpm and gear ratio never change.

Not picking a fight... simply discussing.
Interested to hear your thoughts
 
The torque in an electric motor is immediate and flat. the power increases as the rpm increase.
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That depends on the type and design of electric motor. For cars the torque is typically flat at the lower end, while at the upper end the power is flat.

DDC1D8DB-FDAD-450C-844F-8449B138C692.png

The actual torque output that an engine makes is measured at the crankshaft/ flywheel of the engine with none of the drivetrain attached and is advertised as such.

Sure, but when driving the engine is connected to the rest of the drivetrain so the frictional losses of the drivetrain applies to engine torque as well, other you’d have one power at the flywheel and another power at the driven wheel and and that is physically impossible for a connected system.

1. You yourself have noted that power at the wheel is manipulated by gearing... first quote of this post... and is what i have said from the beginning.

I’ve clearly repeated that power is NOT changed by gearing.

Power = torque * angular velocity

4Nm at 2rad/s = 8W

Now, apply a gear ratio of 0.5:

4Nm*0.5 at 2rad/s/0.5 =
2Nm * 4rad/s = 8W

The gear ratio of 0.5 halved the torque and doubled the speed. Power remains 8W and did NOT change. That’s N-O-T for not. Do you read me, over?

4*2 = 8
2*4 = 8

2. The second example clearly applies a multiplier to TQ

Indeed, well spotted. That is exactly that the gearbox does.

then adjusts gearing to justify simple arithmetic...

No, the adjusted gearing is the very reason why the torque changed.
 
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1. You yourself have noted that power at the wheel is manipulated by gearing... first quote of this post... and is what i have said from the beginning.
No, that was torque. Torque is not power.
A 2:1 gear ratio will rotate the engine twice for every turn of the drive wheel. That means you get twice as many combustion events per rotation of the driven wheel. Twice the torque, half the rotational speed, at the wheels. Power is the rate at which work is done. If you double the torque and reduce the speed by half you're still doing work at the same rate.
2. The second example clearly applies a multiplier to TQ, then adjusts gearing to justify simple arithmetic... but has no relevance in a engine/transmission/diff/wheel power delivery system.
It's the other way around, changing the gear ratio will affect the torque and speed. The whole point was to show that the power is not multiplied by gearing.
 
Sure, but when driving the engine is connected to the rest of the drivetrain so the frictional losses of the drivetrain applies to engine torque as well, other you’d have one power at the flywheel and another power at the driven wheel and and that is physically impossible for a connected system.

Which would be correct as the engine as shown on an engine dyno measured at the crankshaft at x rpm is putting out x torque and x horsepower.

That same engine then connected to a drivetrain which then has its power measured at the drive wheels representing the amount of power actually delivered to the ground at x rpm is putting out x torque and x horsepower.

All measurements of a chassis dyno are performed at a gear ratio of 1:1 to eliminate the changes in the numbers that could be altered as a result of varied gear ratios.

The x horsepower and x torque numbers of the chassis dyno taken at the drive wheels will ALWAYS BE LOWER than the X horsepower and x torque numbers from the engine dyno taken at the crankshaft because of frictional and resistance losses throughout the system.

The engine would still be capable of producing its original figures without the losses that are directly contributed because of the drivetrain being connected to the engine.

You have not reduced the power the engine is capable of making just changing the load at which the measurements are recorded at. That would be why the loss is contributed to drivetrain loss and not engine loss.
 
4Nm at 2rad/s = 8W
Now, apply a gear ratio of 0.5:
4Nm*0.5 at 2rad/s/0.5 =
2Nm * 4rad/s = 8W

The gear ratio of 0.5 halved the torque and doubled the speed. Power remains 8W and did NOT change. That’s N-O-T for not. Do you read me, over?
4*2 = 8
2*4 = 8
wait a minute... all this time your term "speed" is driven wheel speed... not engine rpm?
 
Which would be correct...
We need to drop it.
They are examining rear wheel speed, thru gearing, and using TQ as a linear function at a static point in time... ignoring engine rpm/acceleration and TQ/HP curves/characteristics.
A 2:1 gear ratio will rotate the engine twice for every turn of the drive wheel. That means you get twice as many combustion events per rotation of the driven wheel. Twice the torque...
Right here we have doubled the rpm, but ignored the fact that... that in and of itself has changed the TQ value in a non-linear amount.
I'm out.
 
Right here we have doubled the rpm, but ignored the fact that... that in and of itself has changed the TQ value in a non-linear amount.
I'm out.
I see the misunderstanding now. When I say double the torque and half the speed I mean at the same rpm in a different gear.
That was only to demonstrate the point that gearing does not multiply power.
 
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