which would be better for a 88 civic hatch siR(horsepower) or ls(torque)?

 

Personally I think this is way to vague of a question. Please tell us what you would like to do with your car and then we can answer this question a lot better. Anyways though a Civic can always use more torque, but then again a civic can also always use more horsepower.

 

depends on what u want
drag 1/4 mile or auto cross racing or highway racing???

first chosse that then u can debate which is better

I got a ls/vtec in my EF

so I got the best of both worlds

LoL

 

there's no versus about it.

please read this article and you will be an expert:

http://vettenet.org/torquehp.html

torque is how much, how far.

horsepower is how much, how far, how fast. until you compare much/far (mass and distance) to time, you can't talk about fast. fast is in relation to time.

imagine if hondas couldn't rev past 6000 like most domestics. they'd have to make as much torque as possible, likely have the biggest engine possible, to increase horsepower because they can't do it by revs.

you can't go wrong building a B18C properly. how much power you car will make will be limited by money anyway, and a bigger engine (meh, H22) CAN have bigger power gains for the same type of upgrade, which will pretty much cost the same for any engine.

 

i plan on mostly highway and drag racing

 

By definition to get from point A to point B the fastest you want to maximize your average power output, regarless of torque output.

If you need to haul trailors or boats, or looking to be pinned in your seat, then go for torque. You can have all the torque in the world, but without creating power, you will go no where.

 

Horsepower is the combination of torque and rpm. The beauty of torque and rpm is that you can convert one into the other...so if one car has more torque and another more power, the one with more power will actually end up having more torque at the wheels.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">Horsepower is the combination of torque and rpm. The beauty of torque and rpm is that you can convert one into the other...so if one car has more torque and another more power, the one with more power will actually end up having more torque at the wheels.</TD></TR></TABLE>

Convert from rpm to torque? Those are two entirely different units of measurement, there is not converting from one to the other. Torque output (torque vs. rpm) is going to be totally dependant on the design of the motor. Knowing torque and engine rpm you can CALCULATE the power at that specific rpm, but you cannot convert from torque to power, as they are two different units of measurement.

And you last statement about make no sense either. You said car A has more torque and car B has more power, but car B actually has more torque$%@$%$$? Makes not sense at all. Take any J-body 4 cylinder and compare it to a B16. B16 has more power but less torque, while J-body has more torque but less power.

 

A trasnmission converts rpm into torque, does it not?

That said, the car with more power will have more torque at the wheels because whatever miniscule torque it might have at the crank have will have been multiplied by the transmission. Example: a b16 with J1 transmission will put down in the neighborhood of 1300-1400lb-ft of torque to the wheels in first gear.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">A trasnmission converts rpm into torque, does it not?</TD></TR></TABLE>

No. A transmission tranmits torque from the crankshaft, multiplys it through gearing and distributes it to the driving wheels.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">That said, the car with more power will have more torque at the wheels . . .</TD></TR></TABLE>

This is not true. Power is a function of torque AND engine speed, not just torque. Now you can reword it as such. If car A has more power at X RPM that car B at X RPM, then car A has more torque at X RPM, but you cannot say a car with more power will have more torque, as that simply goes against basic physics, not to mention there are countless production cars out there that prove your statement wrong.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by BryanPendleton &raquo;</TD></TR><TR><TD CLASS="quote">No. A transmission tranmits torque from the crankshaft, multiplys it through gearing and distributes it to the driving wheels.</TD></TR></TABLE>
.....while at the same time reducing the rpm. It multiplies torque at the expense of rpm, or multiplies rpm at the expense of torque. I don't see how the statement "A transmission converts rpm into torque" is invalid.


<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by BryanPendleton &raquo;</TD></TR><TR><TD CLASS="quote">This is not true. Power is a function of torque AND engine speed, not just torque. Now you can reword it as such. If car A has more power at X RPM that car B at X RPM, then car A has more torque at X RPM, but you cannot say a car with more power will have more torque, as that simply goes against basic physics, not to mention there are countless production cars out there that prove your statement wrong.</TD></TR></TABLE>

I don't think we're arguing about the same thing, because I agree with you on that.
What I'm trying to say is if two identical cars (only difference being their drivetrain) are going at the same speed, and car A is producing more power than car B, car A WILL have more torque at the wheels than car B. This is easily proven by the formula HP=TQ*rpm/5252

This remains true even if car B has 100000 times the torque, at the crank, of car A. It is true regardless of the torque output of the two cars at the crank .

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">
.....while at the same time reducing the rpm. It multiplies torque at the expense of rpm, or multiplies rpm at the expense of torque. I don't see how the statement "A transmission converts rpm into torque" is invalid.</TD></TR></TABLE>

I am disaggreeing with your choice of words, as I think you understand, but your choice of words does not make sense. RPM is revolutions per minute and torque is a measure of the magnitue of a moment. You cannot "convert" from one to the other. The transmission stands between the crankshaft and that axle. The transmission increase or decreases the LEVERAGE or GEARING of the crankshaft output. It converts X RPM to Y RPM through a gearing change. It converts A torque to B torque through a gearing/leverage change. Converting rpm to torque or vice versa simply does not make any since.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">I don't think we're arguing about the same thing, because I agree with you on that. What I'm trying to say is if two identical cars (only difference being their drivetrain) are going at the same speed, and car A is producing more power than car B, car A WILL have more torque at the wheels than car B. This is easily proven by the formula HP=TQ*rpm/5252

This remains true even if car B has 100000 times the torque, at the crank, of car A. It is true regardless of the torque output of the two cars at the crank .</TD></TR></TABLE>


If two cars are identical (only transmission gearing changes) going the same speed (MPH, not RPM), the only way one can be making more power at the wheels at the specific speed (MPH) is with a gearing change. And since the engine outputs are identical then the one making more power (Car A) will naturally be making more torque at that specific speed (MPH), simply because one of the car's engine will be operating at a engine rpm/speed that creates more power and thus more torque. With that said Car B cannot have 100000 times the torque of car A at the crank, because Car A is alread making more power and torque, because it is operating at a better engine speed. You are still not getting your point across, what ever it is.

 

if two identical car was in a 1/4 race and the only difference was hp and torque. the car with the higher hp will win. torque is not a variable in the acceleration function because it does not include the time variable.

 

i read in an article that torque is only good up to 40mph, which is like first gear. after that its all about horse power. you can picture it like racing a mustang. forsure he's gonna take you on first gear becuase of all the torque he gots. but then what happens when he starts lossing torque and horse power in his hi rpm range. if you seen there dyno graphs they start lossing power higher in the rpms becuase of all the torque and no horse. you start pulling on him because your power band gives you the most amount of horse power in the hi rpms when torque isn't needed. starting acceleration depends on torque but hi end is all about horse power. so if you want a drag car and a killer highway car then stay with horse. think f1 cars and pro NA drag cars all their looking for is horse power output. but then like someone else said i have a ls/vtec so i have the best of both worlds.

 

[QUOTE=teckno_geek]i read in an article that torque is only good up to 40mph, which is like first gear. after that its all about horse power. [QUOTE]

Torque is only good when you are not moving, because if you are not moving you are not making any power. When you car is sitting still you need to launch with the most torque (assuming traction), the split second your car begins to move, the name of the game is averaging the maximum possible power output, just as 'rice4life' mentioned. Power is a time derivative. To get from point A to point B the fastest, you ultimately want to maximize your average power output, it does not matter how much torque you make.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by BryanPendleton &raquo;</TD></TR><TR><TD CLASS="quote">I am disaggreeing with your choice of words....</TD></TR></TABLE>

All right, that's cool.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">If two cars are identical (only transmission gearing changes) going the same speed (MPH, not RPM), the only way one can be making more power at the wheels at the specific speed (MPH) is with a gearing change. And since the engine outputs are identical then the one making more power (Car A) will naturally be making more torque at that specific speed (MPH), simply because one of the car's engine will be operating at a engine rpm/speed that creates more power and thus more torque. With that said Car B cannot have 100000 times the torque of car A at the crank, because Car A is alread making more power and torque, because it is operating at a better engine speed. You are still not getting your point across, what ever it is.</TD></TR></TABLE>

The engine of car B could very well be making 100000lb-ft of torque and spinning 1rpm. It will then be making 19hp. Car A could only be making 100lb-ft of torque and spinning at 5000rpm thus making 95hp....less torque and more power. If you run them both through a proper transmission car A ends up with much more torque at the wheels. Thus my original statement and point:

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">...so if one car has more torque and another more power, the one with more power will actually end up having more torque at the wheels.</TD></TR></TABLE>

 

think of it this way, an electric motor can make 100 f/lb torque at zero rpm. if you hold the drive shaft still and turn on the electric motor, you still feel torque. or, when you tighten a bolt down to 50 f/lb, when you hit 50 f/lb you are not turning the bolt any farther but you are still applying torque. so without rpm, you are not going to accelerate any where.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">
The engine of car B could very well be making 100000lb-ft of torque and spinning 1rpm. It will then be making 19hp. Car A could only be making 100lb-ft of torque and spinning at 5000rpm thus making 95hp....less torque and more power. If you run them both through a proper transmission car A ends up with much more torque at the wheels. Thus my original statement and point:

...so if one car has more torque and another more power, the one with more power will actually end up having more torque at the wheels.</TD></TR></TABLE>

Am I the only one whom this does not make one bit of sense? Or the relevance? My guess is its your choice of words again, but I will beat this horse until I get it out of you.

Lets disect your "original" statement: "...so if one car has more torque and another more power, the one with more power will actually end up having more torque at the wheels."

Car A has more torque
Car B has more power
but Car B has more torque the wheels.

Lets start with what torque and what power? Are you talking about torque at the crankshaft or actual brake torque (leveraged torque as measured post transmission gearing) at the wheels? Note that brake torque and dynojet torque at the wheels are not the same, unless the transmission and differtial gearing are 1-to-1. Are you talking power at the crank or wheels. Torque at what engine speed (rpm)? Power at what engine speed (rpm)? Or are these peak values?

 

Haha yeah I thought that's what the problem was. I'm starting to think that what I'm trying to say is so simple and obvious to you that you are looking deeper into it than you should be in an effort to try and make it more complicated.

If:
Car A is currently producing more torque at the crank.
Car B is currently producing more power at the crank.

Then:
Car B will be currenlty producing more torque at the wheels.


Power will be equal at the crank and wheels, as we should assume no drivetrain losses. As for what rpm they are at and all that...lets assume car A is at its peak torque and car B at its peak power. Naturally it follows that car A has to be at an rpm lower then car B. But it really doesn't matter. It all comes down to this:

If:
Car B is currently producing more power at the crank than car A.
There are no drivetrain losses.
The wheels of car A and car B are rotating at the same rpm.

Then:
Car B will currenlty have more torque at the wheels than car A.


You will not find an instance when the is not true.
As for the relevance....this all answers the original question of this thread.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">Haha yeah I thought that's what the problem was. I'm starting to think that what I'm trying to say is so simple and obvious to you that you are looking deeper into it than you should be in an effort to try and make it more complicated.

If:
Car A is currently producing more torque at the crank.
Car B is currently producing more power at the crank.

Then:
Car B will be currenlty producing more torque at the wheels.


Power will be equal at the crank and wheels, as we should assume no drivetrain losses. As for what rpm they are at and all that...lets assume car A is at its peak torque and car B at its peak power. Naturally it follows that car A has to be at an rpm lower then car B. But it really doesn't matter.</TD></TR></TABLE>

So you are assuming 0 drivetrain losses.

You assume:
Car A is at X rpm creating peak torque at the crank, which is more than car B
Car B is at Y rpm creating peak power at the crank, which is more than car A.

You claim Car B MUST be creating more brake torque at the wheels, correct?

Well I disagree, because a simple gearing change will cause that conditional to be false. This conditional will be false with something as simple as a gearing change, not to mention it gives no consideration to the torque and power curves of the motor.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">It all comes down to this:

If:
Car B is currently producing more power at the crank than car A.
There are no drivetrain losses.
The wheels of car A and car B are rotating at the same rpm.

Then:
Car B will currenlty have more torque at the wheels than car A.


You will not find an instance when the is not true.
As for the relevance....this all answers the original question of this thread.</TD></TR></TABLE>

this I will agree with, as this can be mathematically proven, as below.

WTQ = brake torque at wheels
HP = power at crank
S = vehical speed
T = torque at crank
RPM = engine speed
GR = gear ration, including transmission and wheel size

You claim if
WTQ1 &gt; WTQ2 if HP1 &gt; HP2 and S1=S2

We all agree that HP = T*RPM/5252, so T = HP*5252/RPM
WTQ = T*GR, so T = WTQ/GR
S = RPM/GR, so RPM = S*GR

Solving for WTQ, using known is as follows:

WTQ/GR = HP*5252/(S*GR)
GR will cancel out leaving:

WTQ = HP*5252/S

So WTQ1 &gt; WTQ2 equate to:
HP1*5252/S1 &gt; HP2*5252/S2

but S1 and S2 are the same per your assumption, so they cancel, as well as the 5252 yielding

HP1 &gt; HP2

Proving you last statement.

I fail to see the relevence to the original question.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by BryanPendleton &raquo;</TD></TR><TR><TD CLASS="quote">
So you are assuming 0 drivetrain losses.

You assume:
Car A is at X rpm creating peak torque at the crank, which is more than car B
Car B is at Y rpm creating peak power at the crank, which is more than car A.

You claim Car B MUST be creating more brake torque at the wheels, correct?

Well I disagree, because a simple gearing change will cause that conditional to be false. This conditional will be false with something as simple as a gearing change, not to mention it gives no consideration to the torque and power curves of the motor. </TD></TR></TABLE>

Well, yeah...if you change gear it will probably cause one of these to change:

Car A is at X rpm creating peak torque at the crank, which is more than car B
Car B is at Y rpm creating peak power at the crank, which is more than car A.

Thus causing the statement: Car B MUST be creating more brake torque at the wheels, to be false.

But then again if you turn the engine off or forget where the gas pedal is it will cause one of these to change as well. The point is if you have a proper transmission and know how to drive chances are the car with more peak power will be faster than the car with more peak torque.

This is precisely why a b16 equipped car will outaccelerate a b18a equipped car. This is also why generally the car with more power will outaccelerate the car with more torque (but less power). There's the direct answer to the original question, and all this talk is the explanation.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">
This is also why generally the car with more power will outaccelerate the car with more torque (but less power).
</TD></TR></TABLE>

I have seen many cases where this isn't true.

A car with more torque generally has a stronger powerband, and power all long the RPM range. Where a car that revs really high and makes more power but less torque, is very peaky.

Sure you may be putting more power to the ground, but if its only for a period of 1,000rpm how much good is it going to do you?

This is all kinda pointless hypothetical stuff, as I am sure everyone knows that power/torque is only one factor of many in what determines how fast a car can accelerate.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by StyleTEG &raquo;</TD></TR><TR><TD CLASS="quote">

I have seen many cases where this isn't true.

A car with more torque generally has a stronger powerband, and power all long the RPM range. Where a car that revs really high and makes more power but less torque, is very peaky.

Sure you may be putting more power to the ground, but if its only for a period of 1,000rpm how much good is it going to do you?

This is all kinda pointless hypothetical stuff, as I am sure everyone knows that power/torque is only one factor of many in what determines how fast a car can accelerate.</TD></TR></TABLE>

Maximizing average power, or the area under the power curve in the rpm range in which the car is drive is what its all about.

 

MY SAY ON THE MATTER IS GO B18C AND YOU SHOULD BE FINE. IF I AM CORRECT A B18C IS GSR SPECS. I CAN'T BELIEVE THERE IS SOMEONE ON THE SITE TRYING TO TELL ME THAT A B16 IS A BETTER ENGINE THAT DOES MORE WORK THAN A B18 THAT IS CRAZY.

IN THIS SITUATION I WOULD GO B18C GSR SPECS.

OR GO B18C5 TYPE R ENGINE 190 HORSEPOWER

 

Dude, why you yelling at us?