drivetrain loss????
 

Ok, so everybody knows that the drivetrain takes up power. And everybody has their on idea of how much is lost, right? In a manual car somewhere around 15-18% and in an auto closer to 25%, right?

Here's where I have a problem with that figuring. Say you had a 100 horse motor, you put it into a manual car. you put it on a dyno and the motor makes 82 hp at the wheels, that would be a loss of 18% or 18 horse, correct?

Now take a 200 horse motor, put it into the same manual car. Am I supposed to believe that it take the 200 hp motor more power to turn the drivetrain that the 100 horse motor? With the current firgures the 200 horse motor would make 164hp. Why does it take more power to turn the wheels with a more powerful motor than it would a weaker motor?

I just don't understand this logic. Basing hp loss on percentages is not a bad way to guess at what your motor is making. Being able to work with a dyno service, I'm going to test a few theories and let you know what I come up with, but I think we will all be somewhat surpized to find that the drivetrain loss is a constant figure, let's say 35hp for example, instead of a percentage.

I don't know. maybe I jsut have too much time on my hands, huh?

Later.

hmmm. an aero engineer and i were talking about this exact thing on a dodge stealth forum about 2 weeks ago. let me find a link for ya.

I got into this argument with one of my firends. I'm interested in knowing what the real answer is.

We also go into the argument that if it mathmatically possiable to figure out Torque if you know the RPM and HP. I disagree that it is mathmatically possiable. Why is it that someone with one kind of motor setup have more torque then another person with a different motor setup but they both have the exact same HP? :confused:

I'm going to throw a link to a dyno site that I'm currently working with. Unfortuantely, it's a bunch of f-Body cars, but it's going to help me figure this thing out.

If you scroll down, you can see me making a pull in a Civic. YEA!!

It made 98hp!!! Kick azz!! Later.

The web address is www.danddmobiledynoservices.com. Check'em out and see. I don't know if the link will work or not, but you can just cut and paste if you want. Or not, whatever you think is best. Ya bunch of Ford freaks!!!!

Later.

Very interesting topic!

Cyberstang: Here are the formulas for hp and torque.


Torque = force * moment arm length

Hp = RPM * Torque / 5252


If two engines make the same hp at the same rpm then their torques will be equal.

If the engines have the same hp but different torques, this would be due to being measured at different rpm's.


Here is a quick example of two machines that have the same hp but different torques. Notice that to do this, there must be a change in rpm.

2000 rpm * 262.6 ft-lbs / 5252 = 100 hp

or

3000 rpm * 175 ft-lbs / 5252 = 100 hp

If rpm's increase, torque has to decrease to keep the same hp.


Hope this helps out a little.

Torque/RPM/HP
 

It is possible to compute torque if you know the HP and the RPM at which that HP is produced. This is because horsepower is a function of torque at a specific rpm. An engine dyno measures the torque that the engine puts out at a certain rpm and then the hp is computed from that torque measurement.

The formula is:

HP = (torque X RPM) / 5252

or

Torque = (HP x 5252) / RPM

Rev

Well Hi Fellas!
 

Hi fellas...

Havn't been on for a while. Just wanted to pop in and this one of course caught my attention...

Here is a nice nerd link to review.... Car Math....

http://www.speeddemonmotorsports.com...trainloss.html

Hope everyone is doing good...

~Jenn~
aka Jane of all Trades

Good post Jenn :)

Hethj7 and Rev:
Ok, after my friend and I had this argument I went and asked my college physics professor how to calculate HP. He told me that HP is a how fast you can move an object with a certin amount of weight in a certin amount of time over a certin amount of distance. So weight, time, and distance is what is needed to calculate distance. HP is a type of power/energy and Torque is a type of force. Now correct me if I'm wrong (or my college professor is wrong) about this HP vs. Torque thing. Or are we arguing about the same thing and I just don't see how this forumla works?... :confused:

Cyberstang -

I think you got it there. The formulas we gave you for hp involved weight (weight is a force and is accounted for when you calculate torque), distance (basically accounted for by rpm's, the revolutions being the distance so to say), and time (again from rpm....time being the per minute part).

If it makes it easier for you to visualize, here is the hp formula substituting force * distance for torque.

HP = Rev. / Min. * Force * Distance / 5252

Remember that the force here is a weight.

Don't know if this helps anymore.....maybe write it out on paper and play with a few numbers like I did in the example in my previous post.

ok, got another question for ya...
so with this formula you are saying that Torque and HP are directly porportional to each other. You can have a chart with both HP and Torque numbers on it. You will be able to look at it and find how much HP you have and it will tell you exactly how much Torque you have? If that is the case, then why do people have different power bands and torque curves when they dyno (some look like a mountain, some is a strait line then falls off at the end, and others have nothing but a strait line across the page)?

Torque/HP
 

I think your physics proffessor is correct. Torque is measured in Lbs./ft. This accounts for force (lbs.) and distance (ft). The RPM accounts for time (revolutions per minute).

The torque curves that you mentioned are largely determined by the cam shaft duration, lift, lobe separation etc. For any of those different torque curves you mentioned there will be a corresponding HP figure for any given RPM. Use that formula HP=(torque X RPM)/5252 and that will be the HP for that spot on the torque curve.

Rev

Sorry for being so difficult about this, it's starting to gel a little more...

Ok, explain this....
I know magazines are not all that dependable when it comes to testing performance parts so I'm not expecting this example to fly very far. Let's say they are testing cat-backs. If they test part A and dyno the car they claim they got 10 more HP and 15 more ft/lbs of Torque. If they test part B on the same exact car they will get the same 10 HP but only put out 12 ft/lbs. If this is true, then the formula would be incorrect seeing as how your using the same cam, on the same car.

Also, at what point do you start to make more HP then torque? When I had my stock motor I pulled a 240HP/300Tq at the wheels. Some of the bigger motors will have more HP then Tq. Why all of the sudden the loss of torque? John Force doesn't have 6000HP/8000ft/lbs. It's more like 6000HP/2000ft/lbs.

Rev is right here.

Also remember that hp is proportional to both torque and rpm. That is another way to explain the differing curves.

Picture a relatively straight torque curve. Its corresponding hp curve will rise all the way through the rpm range. This is because as the rpm's increase and torque stays relatively constant, the hp curve will begin to rise. HP is related to rpm's also, not just the torque.

Yes, that would explain why at stock Mustang weighing at 3300lbs runs a 1/4 mile at 14 seconds but if the same car runs the same 1/4 mile track but ET's at 16seconds it would show that the car did not reach it's peak RPM and/or red line. :D

I must have been posting at the same time as you on my last reply.


Let's look at your question about cat-backs.

If part A gets 10 more hp and 15 more ft-lbs of torque and part b gets 10 more hp and 12 more ft-lbs of torque, how could this happen? Realize that you have left out one of the variables, namely rpm's.

Both parts could give a 10 hp increase but a different torque increase, but these numbers would be rated at different rpm's. To fully analyze these gains, keep in mind that you need to know all 3 variables; hp, torque, and rpm's.

Sorry if I am not explaing this too well. I wish we could sit down together with a dyno chart in front of us.....that would make things much easier :)!

Yeah, I think it's starting to make sense. It would be easier to have charts infront of me. Man, I am starting to feel stupid that I'm not catching on to this easly...:confused: :rolleyes: :(

Gragh overlays
 

One thing the mags do in comparisons, is to overlay the before and after graphs. The numbers they quote are the biggest differences in torque or HP throughout the test, ie. the most difference that can be measured at any given RPM. That emphasizes the benifit of the mod.

Rev

rev - you said want I was trying to say so simply....nice work :D!

Simlest terms
 

We're all just trying to get this into the simplest terms possible and usable.

Rev

I just I had a thought as to the original question in this thread. Regarding whether drivetrain loss should be a percentage or a constant.....

If the losses in the drivetrain are caused by friction, let's go back to what friction is.

Friction = coefficient of friction X Normal Force

If a 200 hp motor is making more force than a 100 hp motor, then the 200 hp motor also creates more friction as a result of this. Therefore, as power increases, so does the friction and the losses in the drivetrain.

In this case, it would seem that using a percentage would be correct. Maybe the question to answer is whether this percentage should remain the same or increase/decrease with more power?

I need to quit wasting all my thinking power on this thread...I have a test tomorrow I will need it for :p !

25%
 

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

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.

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.

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.

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.

Other Variables
 

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

Different dyno operators
 

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

Thanks 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

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.

Incremental Percentage
 

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