The advantages of a NA engine?

[MVF4Rrider]

Originally Posted by Brosef

You're misinterpreting what dodge said. He said that all else equal, lighter wheels will increase power to the ground. Not power at the crank, but power to the ground (i.e. after all drivetrain losses). Wheel/tire rotating mass is a obviously component of drivetrain loss, so what he said was correct.

I wasn't correcting him, just adding to it.

[swamp2]

Originally Posted by Dodge2Dub

It does make sense and I understand the example of adding a passenger vs. being a lonely driver...but...if you change the mass of an object (wheel) that you want to rotate (we'll say decrease it) against a surface, then it becomes easier to rotate it (less force). So, doesn't it stand to reason that if less force is needed to rotate that object ,the translation of applying the same input power is that the object (wheel) will spin faster (increased acceleration)? If so, then if we draw a line on the ground, the power measured when the vehicle crosses over that line with the heavier rotating object (wheel) will be less than the power measured when the vehicle crosses over that line with the lighter object (wheel). Essentially, you are at a higher RPM when you cross the line with the lighter wheel (screw the object reference) then you were with the heavier wheel.

You are going around in a circle here. I can see why you might think this way. There are subtleties here. Just think back to the simple case of a "block" moving on a frictionless surface under some force (not a rotating drive wheel, just a simple directed constant force for simplicity). You must distinguish between the force and the mass. Changing one does not change the other but BOTH can change the acceleration.

Just because something accelerates faster does not imply it must have a larger force applied to it.

Originally Posted by Dodge2Dub

Are the e90 meets in Clairemont on Wednesdays? If so, I never know which Wednesday is a meet day (I believe it's every other Wednesday).

I don't go often either but they are something like every other week. I can't seem to locate the thread on it with any recent activity... I would like to go to one again sometime soon.

[Dodge2Dub]

Originally Posted by swamp2

You are going around in a circle here. I can see why you might think this way. There are subtleties here. Just think back to the simple case of a "block" moving on a frictionless surface under some force (not a rotating drive wheel, just a simple directed constant force for simplicity). You must distinguish between the force and the mass. Changing one does not change the other but BOTH can change the acceleration.

I have to disagree. Changing the mass has a direct correlation to the amount of force needed to move an object to a given point. If the same force is applied for a lightened object, it will arrive at that given point sooner.

In a vehicle with lighter wheels vs. a heavier set, when it crosses the "line" at some distance ahead of the initial take-off position, it will have built more RPM's due to the same force (Tq) being applied to a lighter rotating object. If the power to the ground is measured exactly when the two vehicles cross this "line" then we will see that the vehicle with the lighter wheels has built up more power than the one with the heavier set. I understand that the engine is not making more power, but the relative power to the ground has increased.

[swamp2]

Originally Posted by Dodge2Dub

I have to disagree. Changing the mass has a direct correlation to the amount of force needed to move an object to a given point. If the same force is applied for a lightened object, it will arrive at that given point sooner.

You have a mixture here of correct and incorrect but in essense you are correct. I sort of do as well. In my prior post where I said,

Originally Posted by swamp2

In short when you change a wheel weight (or any drivetrain inertia) you affect performance (acceleration) but you do not affect the power.

I should have been more precise. Specifically, in a vehicle with no engine changes, changing the mass or inertia of the vehicles components does not affect the vehicles peak engine power. However in a race the lighter car (or one with less drivetrain inertia) will obviously accelerate better and hence at any given time (while in the same gear!) it will be producing more instantaneous power. The faster car will end up shifting sooner (same speed but sooner) and just after the shift the heavier car, for a short duration, will be producing substantially more power than the faster/lighter car.

Now onto yours...Your first sentence above is completely incorrect but the second sentence is correct. Any force on any mass makes it move unless overcome by friction or other non conservative forces. For example an any can push a frictionless locomotive, it is just that its acceleration will be incredibly small. I think what you were trying to say is something along these lines: for a given level of acceleration a change in the accelerated mass requires a corresponding change in the force.

Originally Posted by Dodge2Dub

In a vehicle with lighter wheels vs. a heavier set, when it crosses the "line" at some distance ahead of the initial take-off position, it will have built more RPM's due to the same force (Tq) being applied to a lighter rotating object. If the power to the ground is measured exactly when the two vehicles cross this "line" then we will see that the vehicle with the lighter wheels has built up more power than the one with the heavier set. I understand that the engine is not making more power, but the relative power to the ground has increased.

The first part here is also correct upon covering a fixed distance the lightened or lower inertia car will be producing more power when crossing the line. However, it will also be producing more power at almost all times during the run (again just not during times associated with the changed shift times). Power is not built up, energy is what is built up by forces and power. Power is an instantaneous concept (just like speed).

[swamp2]

Originally Posted by Brosef

You're misinterpreting what dodge said. He said that all else equal, lighter wheels will increase power to the ground. Not power at the crank, but power to the ground (i.e. after all drivetrain losses). Wheel/tire rotating mass is a obviously component of drivetrain loss, so what he said was correct.

Again that is completely incorrect. Masses and intertias do not count as losses in any equations. It is incorrect to say that either a heavier car or one with heavier wheels has more drivetrain loss and thus less power to the wheels. It only produces more instantaneous power because it is lighter and thus naturally accelerates better also allowing the engine to reach a higher rpm at any given time and thus more instantaneous power.

For a review the losses on a car are truly losses you can not get that energy back. Masses and rotating components only store energy (during acceleration). Yes their energy is lost say when braking but that is not relevant during our hypothetical drag races. Losses are:

-Tire losses
-Drivetrain losses (transmission, rear end, axles). These are frictional losses that convert power into heat and them it is "lost".
-Aerodynamics. Really aerodynamics enters the equations differently. Aerodynamics simply produce a counteracting force to the drive force that acts more or less across the entire body of the car. But you can also describe this as a loss - the air is actually heated. Just multiply the aero force by the vehicle speed and you get the aero "power loss".
-Auxiliaries: AC, steering pump, water pump, etc. Some might be part of the cars power spec some might not be.
-Clutch: In a MT these only are during the shifts.

All of the losses above except tires and aero go into the overall drive train efficiency and the figure is multiplied by the engine crank torque and divided by wheel radius and vehicle mass to get the primary net drive force. Here the "loss" it typically around 15% and hence the efficiency is 85%. Again these values vary by both engine speed and gear but the typically reported percentages are the peak values at redline.

Hope this helps. It is not trivial at all.

[Dodge2Dub]

Originally Posted by swamp2

Again that is completely incorrect. Masses and intertias do not count as losses in any equations. It is incorrect to say that either a heavier car or one with heavier wheels has more drivetrain loss and thus less power to the wheels. It only produces more instantaneous power because it is lighter and thus naturally accelerates better also allowing the engine to reach a higher rpm at any given time and thus more instantaneous power.

For a review the losses on a car are truly losses you can not get that energy back. Masses and rotating components only store energy (during acceleration). Yes their energy is lost say when braking but that is not relevant during our hypothetical drag races. Losses are:

-Tire losses
-Drivetrain losses (transmission, rear end, axles). These are frictional losses that convert power into heat and them it is "lost".
-Aerodynamics. Really aerodynamics enters the equations differently. Aerodynamics simply produce a counteracting force to the drive force that acts more or less across the entire body of the car. But you can also describe this as a loss - the air is actually heated. Just multiply the aero force by the vehicle speed and you get the aero "power loss".
-Auxiliaries: AC, steering pump, water pump, etc. Some might be part of the cars power spec some might not be.
-Clutch: In a MT these only are during the shifts.

All of the losses above except tires and aero go into the overall drive train efficiency and the figure is multiplied by the engine crank torque and divided by wheel radius and vehicle mass to get the primary net drive force. Here the "loss" it typically around 15% and hence the efficiency is 85%. Again these values vary by both engine speed and gear but the typically reported percentages are the peak values at redline.

Hope this helps. It is not trivial at all.

I think that we are saying the same thing. Lighter = faster...everything else being equal (simple man's concept)! However, I appreciate the feedback and hope that others see the value in the arguments that have been posted thus far.

[swamp2]

Originally Posted by Dodge2Dub

I think that we are saying the same thing. Lighter = faster...everything else being equal (simple man's concept)! However, I appreciate the feedback and hope that others see the value in the arguments that have been posted thus far.

Cheers, it was a good an valuable discussion . I always learn and find new ways to think about the "right" or "best" way to relate automotive performance back to the relevant physics.