Ulfs Power Test: Performance & Review

First of all, congratulations, it's really well done.
I might have a few more hundredths:

1.
Air resistance depends on the air density, which is influenced by pressure and temperature. --> Correction factor for air resistance.

p[mbar]/1013 * 293/(T[°]+273)


2.
The performance is calculated using the formula: square root of (293 / (T + 273)).
It seems that below a certain temperature (around 16°C), the performance remains consistent (with sufficient air mass to achieve the rated torque), but above approximately 16°C, the power decreases with the air density.
How about a temperature correction factor in Excel that looks like this:

=IF(E10<16, 1, 289/(273+E10))


3.
And then there's something a bit more complicated .
You don't just need to accelerate the mass in a straight line; you also need to impart rotational acceleration to things like wheels, shafts, gears, the flywheel, the crankshaft, etc. Of course, this doesn't make as much of a difference as it does with a moped. (With something like that, you can improve acceleration by using a lighter flywheel without increasing the power output.)

a = F / (m + (I * i^2) / r_tires^2)

and the moment of inertia (I) of a rotating mass is m*r^2/2.
Everything located before the gearbox is more important because its effect is squared when calculating the overall transmission ratio (i).

'If anyone knows the weight and diameter of the crankshaft, flywheel, and all other rotating parts, they can post it. (The information for the 1.9L TDI engine is probably of most interest in this forum.)'
I'm not sure if the effort is worth it. The reduction in rotational inertia would need to have the same effect as approximately 20 kg of extra weight when accelerating in a straight line .



I tried the three changes in the benchmark test, and in extreme cases, they make a difference of a few tenths of a second.
I hope I'm going in the right direction.

Ernst S. wrote:

Air resistance depends on the air density, which is influenced by pressure and temperature. --> Correction factor for air resistance. p[mbar]/1013 * 293/(T[°]+273)

Okay, great. Could you please put it into a table and send it to Rainer as an update?

Quote:

Then I can also simply "use" an update without having to create it myself beforehand.

Quote:

2. The performance is calculated using the formula: square root of (293 / (T + 273)). It seems that below a certain temperature (around 16°C), the performance remains consistent (with sufficient air mass to achieve the rated torque), but above approximately 16°C, the power decreases with the air density. How about a temperature correction factor in Excel that looks like this: =IF(E10<16;1;289/(273+E10)) I took the correction factor from a Bosch calculation sheet for dynamometer measurements. My observations (during the 2000-4000 rpm test) are more like a performance variation between approximately -10 and +30 degrees, without a fixed threshold like yours, following the principle of "only from..." Therefore, I prefer to stick to the "official" calculations. 3. And then there's something a bit more complicated

.
You don't just have to accelerate the mass in a straight line; you also have to impart rotational acceleration to things like the wheels, shafts, gears, flywheel, crankshaft, etc. Of course, it doesn't make as much of a difference as it does with a moped. (With something like that, you can improve acceleration by using a lighter flywheel without increasing the power.)

a = F / (m + (I * i^2) / r_tires^2)

and the moment of inertia (I) of a rotating mass is m*r^2/2.
Everything located before the gearbox is more important because its effect is squared when calculating the overall transmission ratio (i).

"If anyone knows the weight and diameter of the crankshaft, flywheel, and all other rotating parts, they can post it. (The information for the 1.9L TDI engine is probably of most interest in this forum.)"
I'm not sure if the effort is worth it. The reduction in rotational inertia would need to have the same effect as approximately 20 kg of extra weight when accelerating in a straight line }.

"Hehe, that's definitely advanced math. Especially because the whole thing is still dependent on the gear ratio (the lower the gear, the more the crank mechanism and flywheel dominate)."
I have to laugh out loud }.
I suggest incorporating a thumb rule as an addition to the vehicle weight, which could be relatively easily implemented – provided we have a reasonable estimate for the amount of the addition.
The thumb value could, of course, be adjusted based on speed to get even closer to reality...

I will incorporate the air resistance factor, along with pressure and temperature. If I have time, I will also investigate how the density of the air changes with humidity, or whether that can be neglected.


Regarding the second point, I wasn't sure about it anyway.
I just thought there would be a maximum injection amount. But it's quite possible that it's only limited by a specific lambda value (soot limit). And the more air the mass airflow sensor (MAF) indicates, the more fuel is injected.
But unfortunately, I don't know any more details about that, so please stick with your Bosch .



Regarding the moment of inertia: I will modify the acceleration calculation and provide an estimated moment of inertia. Then, for example, people with all-wheel drive can increase this value due to the additional rotating masses. (They'll have to compensate for the loss by increasing the output significantly anyway).


I hope all of this works out. For comparison, a tip: On the VW website (or in any car manual), you can find the acceleration figures from 0-80 km/h and from 0-100 km/h.
Simply compare the difference with the time interval of 80-100 from your benchmark test. It should also be in the correct gear.
And with a little bit of moment of inertia, you're already very close to the values specified by VW .

So, humidity definitely has no effect on air resistance when it's calculated based on total pressure and temperature.

The engine power should actually be affected by this... The partial pressure of the air will be lower... During combustion, the enthalpy of the water vapor must be increased... But I'm going to stop analyzing this now .