Re: 5th gear upgrade. .70 ratio!
When did I say anything at all what-so-ever about using the max torque of the engine at 1900??? You are misunderstanding my posts, go and re-read them before you continue to spout off with stupid acusations of WOT with the handbrake applied to maintain 1900 RPM.
Quick rewind to David594's post on 08 Oct.
The main advantage in terms of economy should be the fact that you are keeping the car closer to its max torque. And the max torque should also be the where the car is nearest its maximum mechanical effeciency. This is what your gains would be from, not friction losses.
This was followed by Jackbombay on 09 Oct
Lug Nut, I think you misunderstood the poster you quoted. He, IMO, is saying that driving at peak torque RPM, regardless of right foot position will make HP more efficiently than any other RPM, thus the gains from this mod are more than lower friction, but also more efficient production of any given amount of HP due to operating closer to the peak torque output RPM.
It was in your paraphrase of David that you made his post appear to say that driving at 1900 rpm is most efficient.
David was talking about driving at maximum torque, but there was no reference to any specified engine rpm. I interpreted this as being the maximum torque at any given engine speed.
You then came out with the "peak torque output RPM" implying 1900 rpm. My facetious comment about driving with the brakes on would definitely bring the torque requirement up closer to the maximum available and would satisfy the conditions you paraphrased David as saying are needed for most efficient operation.
If I mis-interpreted your paraphrase, then I apologise.
Horsepower is defined mathematically lb*ft of torque times revolutions per minute, all over the constant of 5252. Out TDI engines make power more efficiently (whether mpg or g/hp/hr) at a higher percentage of possible torque, not necessarily at whichever gear brings the engine rpm closer to 1900 rpm. The specific fuel consumption rate of operating as some horsepower produced with the engine at 1500 rpm will be less than the fuel consumption rate of operating at that same horsepower but with the engine at 1900 rpm. The greater percentage of the lesser maximum torque possible at 1500 rpm is more efficient than a smaller percent of the greater maximum torque possible at 1900 rpm. In short, the engine will be more efficient driving the car at 40 mph in fifth gear at 1500 than it will in fourth at 1900. That is whether the miles per gallon are considered or if the specific fuel per unit of power is considered.
If you wish to now say that the torque produced is higher at 40 mph in 5th gear than at 40 mph in 4th gear based solely on the proportional decrease in rpm and that justifies a 'more torque is more efficient' claim, you may.
I still hold that the increase in efficiency is due to the required torque being a greater percent of the maximum torque possible at that rpm, rather than merely being a bigger number.
If you feel I still don't fully comprehend what you have said, or if you think I'm unfairly attributing statements to you, you may freely reply here or in a private message.