drivetrain loss????

[Rev]

Well I'm saying 25 % because it's a nice round figure. I can do 25% loss in my head , so it's pretty damned convenient. Yes, yes, 25% loss sounds better all the time. I'm sure of it now, LOL.

Rev

[bigwhitecobra]

Well, I think we all know that I am not a genius, nor a physics major, but it seems like the more power you have the lower the loss would be. Think of it as pushing a car. With one person pushing it, it doesn't move very well. But with two people, it moves better and easier. Double the power and the load stays the same, the more work can be done, as in pushing the car. I may be going in the wrong direction, but this is how I see it. Later.

[Hethj7]

Originally posted by bigwhitecobra
Double the power and the load stays the same, the more work can be done, as in pushing the car. I may be going in the wrong direction, but this is how I see it. Later.

That is my point. In an engine the load does not stay the same. The friction is providing the load in the drivetrain, and friction is directly proportional to to power (force). Therefore, more power gives more friction, which leads to a larger load that must be overcome.

Again, friction = coefficient of friction X the force normal to the surface. A greater force will give more friction to overcome.


I could be wrong in that this doesn't apply in an engine, but it is my theory.

[mustangman65_79]

Well, I was thinking about the part where, a manual uses 15% loss(round fiqure) no matter what, and here's what I gather about it. The faster u try to spin something, lets say, a merry-go-round. If three ppl are on, and one's pushing, it's hard, and takes a while to go faster. Now, lets say three ppl are on, and three are pushing, it seems easier to push, but colectivly, your still losing about 15%. Yes, it went turned a higher speed, faster, but the weight still needs the same amount of power to turn it. The faster u want it to turn, the more power it takes.

I hope this kinda helped. I don't know if I get the point across or not.

[bigwhitecobra]

Well, I was thinking about the part where, a manual uses 15% loss(round fiqure) no matter what, and here's what I gather about it. The faster u try to spin something, lets say, a merry-go-round. If three ppl are on, and one's pushing, it's hard, and takes a while to go faster. Now, lets say three ppl are on, and three are pushing, it seems easier to push, but colectivly, your still losing about 15%. Yes, it went turned a higher speed, faster, but the weight still needs the same amount of power to turn it. The faster u want it to turn, the more power it takes.

I think we are saying the same thing. I need to think about this some more so that I can formulate the best way to say what I am thinking.


My head hurts. Later.

IMO........we are stil missing part of the real life application equation. Human skill/reaction time/added weight. Unless you can totally cumputerize the aspect of acceleration on a dyno/chassis dyno, you must take into account that a human person pressing an accelerator pedal will produce a variable each time. This is true for drag racing and 1/4 mile times. Human reaction time to the "tree" changing and the reactionary time it takes the brain to engage the foot on the accelerator.

So, again...unless you are working a computer model with fed calculations and variables, or you have a perfect controlled environmental study, there are variables that you are not taking into account in the equation that was give by your professor.

I love this thread guys...LOL

Jenn

[Rev]

Very true Jane. And different dyno opperators or different dynos (even the same brand and model) will deliver different numbers. Not to mention different atmospheric conditions from one run to another. The correction programs for atmospheric conditions are fairly crude also.

Rev

It is important for all of us to keep the fundamental application of the science and engineering in mind when reading articles and popping around math formulas. It is not so easy to compare a manual drivetrain with an automatic and then calcualte which loses more power. There are issues with oil viscosities that change the formula and weight added by driver, reaction time, tires, tread patterns, environmental conditions.

So, in the end..there is no one simple answer. You can make a pretty good perfect hypothesis, but once you move it out into the "real world"...things change...

Good Luck Fellas and nice way to discuss math and cars....Ahhhhh..good to be back!

Jenn

[Hethj7]

Yea, I understand that there are many more variables than just "friction" within the drivetrain. We all know that the real world applications are quite different than those found in a book.

However, regardless of some of these other variables, such as driver or dyno operator, the basic question of whether drivetrain loss is a constant or a percentage should be able to be examined.

My reasoning stating that friction increases with force is just a stepping stone and obviously leaves some questions to be asked. But, I think it is good start as to why the loss may not be a constant, but rather a percentage. I am not saying it is right, but it seems tobe heading in the right direction.

Hethj7

Based on the "virgin" math..It is an incremental percentage then. As the RPMs increase, the TQ figures encounter drivetrain loss due to friction and resistance.

http://www.charm.net/~mchaney/dynoruns/chevford.htm

Here is a good "home" graph that shows you the #s....

Enjoy...This is a great thread fellas!

Jenn