Drivetrain Loss
#1
05-08-2004 | 02:02 PM
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Posts: n/a
Drivetrain Loss
Ok the engineer in me doesn't understand something...Why is it that drivetrain loss is always expressed as a percentage? We all know a MT to lose 17% +/- regardless of the hp rating. Lets just say for the sake of arguement, I have the same drivetrain for a 1000 hp and 280 hp engine. This suggests that I lose 170 with the 1000 hp engine and only 45 with the 280 Hp engine. Makes no sense to me. As I understand it the losses are due to friction. If that were the case wouldn't you just lose a set amount of Hp through the drivetrain regardless of engine output?
Give me your thoughts.
Sean
G35s 6MT Garnet Fire, Premium, Aero, Aero Body Kit, Winter Sport
Give me your thoughts.
Sean
G35s 6MT Garnet Fire, Premium, Aero, Aero Body Kit, Winter Sport
#2
05-08-2004 | 03:46 PM
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Joined: Dec 2003
Posts: 2,514
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From: Marietta, Georgia
Re: Drivetrain Loss
Obviously the percentage varies with tires, tire temperatures, fluid temperatures, etc even hold down force on the straps.
It really only applies to peak rpms............what a group of testers found was necessary to convert some average RWHP to the published numbers given by manufacturers.
The power losses must be translated to heat rise - in automatics the greatest rise is in the ATF........in manuals the tires , then the diff fluid.
After all what counts is the RWHP not the flywheel power.
It really only applies to peak rpms............what a group of testers found was necessary to convert some average RWHP to the published numbers given by manufacturers.
The power losses must be translated to heat rise - in automatics the greatest rise is in the ATF........in manuals the tires , then the diff fluid.
After all what counts is the RWHP not the flywheel power.
#3
05-11-2004 | 04:51 AM
Re: Drivetrain Loss
well, i'm no engineer, so i don't even think i should attempt to, well........anywho, isn't friction just a function of force? this force acts against the force put on it, so if there is more force, there is more friction, yes? there is a point when the initial force overcomes friction, hence movement. i hope that kind of brings things on a new light, instead of either confusing the hell out of you or making me look like a, well, nevermind......
n1cK
jUs' bE [img]/w3timages/icons/cool.gif[/img]
n1cK
jUs' bE [img]/w3timages/icons/cool.gif[/img]
#6
05-11-2004 | 07:29 PM
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From: LA
#7
05-11-2004 | 08:47 PM
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Joined: Feb 2004
Posts: 550
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Re: Drivetrain Loss
<blockquote><font class="small">In reply to:</font><hr>
well, i'm no engineer, so i don't even think i should attempt to, well........anywho, isn't friction just a function of force? this force acts against the force put on it, so if there is more force, there is more friction, yes? there is a point when the initial force overcomes friction, hence movement. i hope that kind of brings things on a new light, instead of either confusing the hell out of you or making me look like a, well, nevermind......
<hr></blockquote>
This is definitly in the right direction=). There are actually 2 types of friction forces (static and kinetic). To picture this think of looking through a microscope at 2 different surfaces pushed together (like a box on a table). Can you imagine how minature "valleys" of the table would be filled by little peaks of the box? To get the box sliding, a force would have to be applied so that those peaks would be pushed out of the valleys. This force is the static friction. The kinetic friction comes into play when the box is already moving. Those peaks would still try to come to rest within the valleys; however, because the box is moving it requires less force to overcome this tendancy. This is why if you have ever pushed a very heavy box across a table (or floor) it seems to take more force to get it started then to have it continuously sliding.
So back to the origional question. To really simplify just look at the most basic kinetic friction formula:
Fk (kinetic friction) = u (coefficient of friction) * Fn (normal force).
The coefficient of friction is a constant value (<1). So essentially it is a constant percentage just usually expressed in decimal form. =)
"I'm not saying you're full of BS, per se" - Zimbo
-JustICE
well, i'm no engineer, so i don't even think i should attempt to, well........anywho, isn't friction just a function of force? this force acts against the force put on it, so if there is more force, there is more friction, yes? there is a point when the initial force overcomes friction, hence movement. i hope that kind of brings things on a new light, instead of either confusing the hell out of you or making me look like a, well, nevermind......
<hr></blockquote>
This is definitly in the right direction=). There are actually 2 types of friction forces (static and kinetic). To picture this think of looking through a microscope at 2 different surfaces pushed together (like a box on a table). Can you imagine how minature "valleys" of the table would be filled by little peaks of the box? To get the box sliding, a force would have to be applied so that those peaks would be pushed out of the valleys. This force is the static friction. The kinetic friction comes into play when the box is already moving. Those peaks would still try to come to rest within the valleys; however, because the box is moving it requires less force to overcome this tendancy. This is why if you have ever pushed a very heavy box across a table (or floor) it seems to take more force to get it started then to have it continuously sliding.
So back to the origional question. To really simplify just look at the most basic kinetic friction formula:
Fk (kinetic friction) = u (coefficient of friction) * Fn (normal force).
The coefficient of friction is a constant value (<1). So essentially it is a constant percentage just usually expressed in decimal form. =)
"I'm not saying you're full of BS, per se" - Zimbo
-JustICE
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#8
05-12-2004 | 01:53 AM
Re: Drivetrain Loss
well, i'm glad all are in agreement so far (i guess i'm not so dumb after-all, but i could be speaking too soon 
all in all, i agree with justICE in that his explanation is the reason why drivetrain loss is expressed as a percentage rather than a set value. the coefficient of friction would also tend to differ between different kinds of materials/metals, as well as lubricants used.
n1cK
jUs' bE [img]/w3timages/icons/cool.gif[/img]
all in all, i agree with justICE in that his explanation is the reason why drivetrain loss is expressed as a percentage rather than a set value. the coefficient of friction would also tend to differ between different kinds of materials/metals, as well as lubricants used.n1cK
jUs' bE [img]/w3timages/icons/cool.gif[/img]
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