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View Poll Results: What kind of power loss do you think your car experiances? | |||
15-18% for manuals | 7 | 43.75% | |
20-25% for autos | 4 | 25.00% | |
A constant figure | 2 | 12.50% | |
don't know | 3 | 18.75% | |
don't care, just want to go fast! | 0 | 0% | |
Multiple Choice Poll. Voters: 16. This poll is closed |
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03-05-2003, 09:44 AM | #1 |
Huh? Whatcha said?
Join Date: Oct 2000
Location: Fayetteville, NC
Posts: 1,073
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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.
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2003 GT White---STOLEN May you burn in hell whoever took it. Dumbazz didn't even get the good engine that was in the garage. |
03-05-2003, 02:13 PM | #2 |
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Join Date: Jul 2002
Posts: 100
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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.
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03-05-2003, 03:26 PM | #3 |
347ci of HORSE POWER!!!
Join Date: Jul 2001
Location: Houston, TX
Posts: 1,416
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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? |
03-05-2003, 06:39 PM | #4 |
Huh? Whatcha said?
Join Date: Oct 2000
Location: Fayetteville, NC
Posts: 1,073
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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.
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2003 GT White---STOLEN May you burn in hell whoever took it. Dumbazz didn't even get the good engine that was in the garage. Last edited by bigwhitecobra; 03-05-2003 at 06:48 PM.. |
03-05-2003, 08:35 PM | #5 |
Mizzou Tigers
Join Date: Apr 1999
Location: weston, MO United States
Posts: 1,455
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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.
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2006 Mustang GT 1990 LX GT-40 motor 262 horsepower, 307ft-lbs (sold but forever loved) 1998 Contour SVT Rice Haters Club Member #244 |
03-05-2003, 08:39 PM | #6 | |
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Join Date: Nov 1998
Location: Houston, Tx.
Posts: 3,887
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Torque/RPM/HP
Quote:
The formula is: HP = (torque X RPM) / 5252 or Torque = (HP x 5252) / RPM Rev
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'66 Coupe, 306, 350-375 HP, C-4, 13.07 e.t., 104.8 mph, 1/4 mi. O.B.C. #2 '66 coupe Last edited by Rev; 03-05-2003 at 09:05 PM.. |
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03-05-2003, 08:50 PM | #7 |
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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 |
03-06-2003, 08:29 AM | #8 |
347ci of HORSE POWER!!!
Join Date: Jul 2001
Location: Houston, TX
Posts: 1,416
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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?... |
03-06-2003, 12:19 PM | #9 |
Mizzou Tigers
Join Date: Apr 1999
Location: weston, MO United States
Posts: 1,455
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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.
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2006 Mustang GT 1990 LX GT-40 motor 262 horsepower, 307ft-lbs (sold but forever loved) 1998 Contour SVT Rice Haters Club Member #244 |
03-06-2003, 12:28 PM | #10 |
347ci of HORSE POWER!!!
Join Date: Jul 2001
Location: Houston, TX
Posts: 1,416
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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)? |
03-06-2003, 01:39 PM | #11 |
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Join Date: Nov 1998
Location: Houston, Tx.
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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
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'66 Coupe, 306, 350-375 HP, C-4, 13.07 e.t., 104.8 mph, 1/4 mi. O.B.C. #2 '66 coupe |
03-06-2003, 01:54 PM | #12 |
347ci of HORSE POWER!!!
Join Date: Jul 2001
Location: Houston, TX
Posts: 1,416
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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. |
03-06-2003, 01:55 PM | #13 |
Mizzou Tigers
Join Date: Apr 1999
Location: weston, MO United States
Posts: 1,455
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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.
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2006 Mustang GT 1990 LX GT-40 motor 262 horsepower, 307ft-lbs (sold but forever loved) 1998 Contour SVT Rice Haters Club Member #244 Last edited by Hethj7; 03-06-2003 at 03:36 PM.. |
03-06-2003, 02:01 PM | #14 | |
347ci of HORSE POWER!!!
Join Date: Jul 2001
Location: Houston, TX
Posts: 1,416
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Quote:
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03-06-2003, 03:50 PM | #15 |
Mizzou Tigers
Join Date: Apr 1999
Location: weston, MO United States
Posts: 1,455
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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 !
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2006 Mustang GT 1990 LX GT-40 motor 262 horsepower, 307ft-lbs (sold but forever loved) 1998 Contour SVT Rice Haters Club Member #244 |
03-06-2003, 04:26 PM | #16 |
347ci of HORSE POWER!!!
Join Date: Jul 2001
Location: Houston, TX
Posts: 1,416
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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...
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03-06-2003, 07:11 PM | #17 | |
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Gragh overlays
Quote:
Rev
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'66 Coupe, 306, 350-375 HP, C-4, 13.07 e.t., 104.8 mph, 1/4 mi. O.B.C. #2 '66 coupe |
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03-06-2003, 09:33 PM | #18 |
Mizzou Tigers
Join Date: Apr 1999
Location: weston, MO United States
Posts: 1,455
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rev - you said want I was trying to say so simply....nice work !
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2006 Mustang GT 1990 LX GT-40 motor 262 horsepower, 307ft-lbs (sold but forever loved) 1998 Contour SVT Rice Haters Club Member #244 |
03-07-2003, 12:23 AM | #20 |
Mizzou Tigers
Join Date: Apr 1999
Location: weston, MO United States
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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 !
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2006 Mustang GT 1990 LX GT-40 motor 262 horsepower, 307ft-lbs (sold but forever loved) 1998 Contour SVT Rice Haters Club Member #244 |
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