Engine, Drivetrain & Forced-Induction
Have Technical Questions or Done Modifications to the G37? Find out the answer in here!

How come G37 torque is low?

Thread Tools
 
Search this Thread
 
Old 12-10-2007 | 07:48 PM
  #16  
mal_TX's Avatar
mal_TX
Registered User
 
Joined: Jan 2006
Posts: 547
Likes: 0
Well put Kahboom. I know you and I have discussed gear ratios to death elsewhere but since your post focused so much on the engine's characteristics, I felt the need to reinforce that gearing is VITAL to the performance of the car.

Thus we need to consider all of that together, since you don't drive the engine, you drive the car. It's pretty complex and thus we use things like 0-60, 0-150 and 1/4 mile numbers to guage performance.

I guess what I'm saying in the context of this thread is let's not judge a car by any one component -- especially something like a peak torque number.
Old 12-10-2007 | 07:58 PM
  #17  
shumby's Avatar
shumby
Registered User
iTrader: (3)
 
Joined: Sep 2007
Posts: 2,668
Likes: 5
From: With your mom
^^^ well put
Old 12-10-2007 | 08:40 PM
  #18  
bboysteele's Avatar
bboysteele
Premier Member
 
Joined: Jul 2007
Posts: 3,346
Likes: 3
From: Colorado Springs, CO
Originally Posted by mal_TX
Well put Kahboom. I know you and I have discussed gear ratios to death elsewhere but since your post focused so much on the engine's characteristics, I felt the need to reinforce that gearing is VITAL to the performance of the car.

Thus we need to consider all of that together, since you don't drive the engine, you drive the car. It's pretty complex and thus we use things like 0-60, 0-150 and 1/4 mile numbers to guage performance.

I guess what I'm saying in the context of this thread is let's not judge a car by any one component -- especially something like a peak torque number.
Very true. I love the power in this car especially the passing power at higher speeds. I am just waiting for some mods to come out that can give the car a little more umph at the start without taking away from the top end power. The slower starts are the only thing I don't really like on the car. Not that it is bad but I just want a little more quickness off the line.
Old 12-11-2007 | 12:18 AM
  #19  
KAHBOOM's Avatar
KAHBOOM
Super Moderator of Pwnage
iTrader: (4)
 
Joined: Jun 2003
Posts: 2,693
Likes: 195
From: NC
Originally Posted by mal_TX
Well put Kahboom. I know you and I have discussed gear ratios to death elsewhere but since your post focused so much on the engine's characteristics, I felt the need to reinforce that gearing is VITAL to the performance of the car.

Thus we need to consider all of that together, since you don't drive the engine, you drive the car. It's pretty complex and thus we use things like 0-60, 0-150 and 1/4 mile numbers to guage performance.

I guess what I'm saying in the context of this thread is let's not judge a car by any one component -- especially something like a peak torque number.
so true!
Old 01-03-2009 | 03:17 PM
  #20  
G_Thang's Avatar
G_Thang
Registered User
 
Joined: Dec 2008
Posts: 344
Likes: 2
From: the Big KY
The new VQ revs several hundred rpm's higher than the old VQ. From Power = 1/5252*torque*rpm, you see that power is proportional to rpm. The increase from 298hp (or what ever it was) to 330 is due to the increased redline. Using the real numbers you get: Power = 1/5252*(270ft-lbs*7500rpm) = 385.6hp, which is 55hp more than the G is advertised to make. Maybe 270ft-lbs is indicated torque (the torque produced in the cylinder and before any parasidic losses are applied), but I think its a brake number. I'll have to do some digging and see where the discrepancy is? Can anyone help me out here? I'll drag out some of my old engineering books and see if I can figure this out.
Old 01-03-2009 | 08:58 PM
  #21  
jungwoo0622's Avatar
jungwoo0622
Registered User
 
Joined: May 2008
Posts: 157
Likes: 0
From: los angeles
Originally Posted by G_Thang
The new VQ revs several hundred rpm's higher than the old VQ. From Power = 1/5252*torque*rpm, you see that power is proportional to rpm. The increase from 298hp (or what ever it was) to 330 is due to the increased redline. Using the real numbers you get: Power = 1/5252*(270ft-lbs*7500rpm) = 385.6hp, which is 55hp more than the G is advertised to make. Maybe 270ft-lbs is indicated torque (the torque produced in the cylinder and before any parasidic losses are applied), but I think its a brake number. I'll have to do some digging and see where the discrepancy is? Can anyone help me out here? I'll drag out some of my old engineering books and see if I can figure this out.
torque will be max at the around 5000 so
with 7500rpm torque will be much lower
it will be like this
1/5252*2??ft-lbs*7500rpm = 330hp
Old 01-03-2009 | 10:47 PM
  #22  
G_Thang's Avatar
G_Thang
Registered User
 
Joined: Dec 2008
Posts: 344
Likes: 2
From: the Big KY
Ah, yeah exactly. I remember this now. Your not taking torque max and redlline numbers. If you look at a lug curve (torque curve) and look at the end of the curve close to redline rpm, you should see that torque is decreasing less than a 45 degree angel. The point at which the curve decreases more than than 45 degrees is your max power. This point is max power because torque is decreasing at a faster rate then rpm is increasing, hence you have reached max power (since power = 1/5252*torque*rpm). Hope this makes sense.
Old 01-04-2009 | 12:44 AM
  #23  
Mike's Avatar
Mike
Registered User
iTrader: (1)
 
Joined: Feb 2008
Posts: 4,549
Likes: 20
Torque is a function of displacement.

1.0L displacement is about 75 ft/lbs of torque on any N/A engine.

Try looking up everything from corollas, civics, accords, M3s, Ferraris, porsches, etc.

Any NA engine will do roughly 75 ft/lbs of torque.

Turbos/superchargers change things.
Old 01-04-2009 | 01:03 AM
  #24  
G_Thang's Avatar
G_Thang
Registered User
 
Joined: Dec 2008
Posts: 344
Likes: 2
From: the Big KY
Your thinking to BMEP or brake mean effictive pressure. BMEP is a ROUGH was to compare apples to apples when considering outputs of (for example) a B16A2 and a VQ37. BMEP is the mean pressure that if applied to the top of a piston from top to bottom (stroke) would produce the given brake torque of the engine. BMEP is only theoretical and DOES NOT represent actual cylinder pressures. Its only a formula used for comparison (like Volumetric Efficiency). Torque is not a function of displacement. Torque = Force x Distance, which basically means the longer the crank throw (distance from the crank journal center to the crank centerline) the more torque you create.
Old 01-08-2009 | 05:53 PM
  #25  
matrix311's Avatar
matrix311
Registered User
 
Joined: Mar 2008
Posts: 32
Likes: 0
I'm pushing 340ft-lb tq in my boosted 4 banger I love torque!
Old 01-08-2009 | 08:03 PM
  #26  
G_Thang's Avatar
G_Thang
Registered User
 
Joined: Dec 2008
Posts: 344
Likes: 2
From: the Big KY
Originally Posted by mal_TX
Hmmm.. torque AT THE WHEELS is acceleration. Since the engine's output is channeled through a transmission that multiplies the torque, the gear ratio becomes important when considering a car's acceleration. The shorter the gear ratio, the more the torque is multiplied and the faster the input has to turn. Since horsepower, as you stated, is directly related to RPM (in fact, describes the engine's relationship of torque to RPM), horsepower is an important component of the acceleration of a car.

Short version: Very high rpm == lots of torque multiplication at the wheels == fast car, given reasonable torque numbers. Real world example: Formula 1 race car, <200lb/ft torque, 19,000rpm redline, 800HP, 0-100-0 in <5s, top speed about 230mph, nurburgring time < 6min.
Torque multiplication is dependant upon gear ration, NOT rpm. If you had a F1 car run wide open at 19,000 this in no way necessarily means its multiplying torque. When you talk about "overdrive" in an transmission what you are really saying it its output shaft is turning faster than its input shaft. In this instance torque is low and IS NOT mutliplied. Torque is ONLY mutliplied when you are running a ratio lower than a 1:1, beyond that output torque is lower than input torque. If your in first gear you have lots of torque (torque multiplication), but no speed. Conversely, if your running in 6th gear you have to down shift to climb a steep hill. You clearly see torque and speed are INVERSELY proportional and torque is independant upon rpm.
Old 01-09-2009 | 02:32 AM
  #27  
mal_TX's Avatar
mal_TX
Registered User
 
Joined: Jan 2006
Posts: 547
Likes: 0
Originally Posted by G_Thang
Torque multiplication is dependant upon gear ration, NOT rpm. If you had a F1 car run wide open at 19,000 this in no way necessarily means its multiplying torque. When you talk about "overdrive" in an transmission what you are really saying it its output shaft is turning faster than its input shaft. In this instance torque is low and IS NOT mutliplied. Torque is ONLY mutliplied when you are running a ratio lower than a 1:1, beyond that output torque is lower than input torque. If your in first gear you have lots of torque (torque multiplication), but no speed. Conversely, if your running in 6th gear you have to down shift to climb a steep hill. You clearly see torque and speed are INVERSELY proportional and torque is independant upon rpm.
LOLs I dunno what size wheels you are thinking of, but when I think about normal cars running engines turning at 19,000 rpm, I think that better be some f-ing short gear ratios!

Point being -- the higher the revs the shorter the gears can be, which means more torque at the wheels through multiplication. While you are (of course) stating facts, unless the car in question is rolling on quarter-sized wheels or is traveling at hundreds (thousands?) of MPH, a high rpm is going to mean a shorter gear ratio is in play. Especially when we are using the example of a 19,000 rpm F1 car.
Old 01-09-2009 | 12:41 PM
  #28  
Mike's Avatar
Mike
Registered User
iTrader: (1)
 
Joined: Feb 2008
Posts: 4,549
Likes: 20
Originally Posted by G_Thang
Your thinking to BMEP or brake mean effictive pressure. BMEP is a ROUGH was to compare apples to apples when considering outputs of (for example) a B16A2 and a VQ37. BMEP is the mean pressure that if applied to the top of a piston from top to bottom (stroke) would produce the given brake torque of the engine. BMEP is only theoretical and DOES NOT represent actual cylinder pressures. Its only a formula used for comparison (like Volumetric Efficiency). Torque is not a function of displacement. Torque = Force x Distance, which basically means the longer the crank throw (distance from the crank journal center to the crank centerline) the more torque you create.
Completely correct. However, I'd like to add that increased surface area (bore) will increase the force.

force = bore, and distance = stroke.

Force x distance = torque

Bore x stroke = displacement.


Very crude, I know, but it does work...
Old 01-09-2009 | 01:59 PM
  #29  
G_Thang's Avatar
G_Thang
Registered User
 
Joined: Dec 2008
Posts: 344
Likes: 2
From: the Big KY
Originally Posted by mal_TX
LOLs I dunno what size wheels you are thinking of, but when I think about normal cars running engines turning at 19,000 rpm, I think that better be some f-ing short gear ratios!

Point being -- the higher the revs the shorter the gears can be, which means more torque at the wheels through multiplication. While you are (of course) stating facts, unless the car in question is rolling on quarter-sized wheels or is traveling at hundreds (thousands?) of MPH, a high rpm is going to mean a shorter gear ratio is in play. Especially when we are using the example of a 19,000 rpm F1 car.
My point still remains... Torque IS NOT directly dependent upon rpm then as you just said. A consequence low ratios is high rpm's to obtain reasonable speeds, so yeah crudely it could be viewed in this manner. If your speaking of engineering mechanics and physics (which I am), there IS NO direct relation. Also.. How often do you see a "Normal car" churning out 19,000 rpm????

The reason an F1 cars engines is built to turn those kinds of rpm's is because the cars travel road courses and are constantly in the corners, and having such a high rpm allows for a super broad power band, thus the need for shifting is greatly reduced.
Old 01-09-2009 | 02:09 PM
  #30  
DetroitG37Joe's Avatar
DetroitG37Joe
Registered User
iTrader: (2)
 
Joined: Oct 2007
Posts: 2,335
Likes: 1
From: Metro Detroit, MI [Sterling Hts.]
This is a great thread, shows how useful myg37.com can be - simple question that could have been blown off with quick answers was instead delved into with a lot of thought and educated speculations. Much thanks to the "Kahboom & Mal_TX" fountain of knowledge. I know I learned a lot. Thanks guyz!


Quick Reply: How come G37 torque is low?



All times are GMT -4. The time now is 01:20 PM.