<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nightryder21 &raquo;</TD></TR><TR><TD CLASS="quote">... Could you say "Machine B" has more torque than "Machine A"? If so, how can we get our engines to act more like "Machine B"?</TD></TR></TABLE>Machine B produces more torque than machine A. Piston engines make torque thru the pistons' downward force acting on the crank. The only way to increase torque is to increase the displacement of the engine, or increase the pressure inside the cylinders. Go back to my earlier post.

If you include the transmission within your concept of 'machine', then you can increase torque by lowering the gear ratio. So your description of A & B might actually be the SAME motor with different gearboxes. I've actually got 3/8" and 1/2" drills that look like the same motor with different gearboxes, so that's a good example.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by 4crx4me &raquo;</TD></TR><TR><TD CLASS="quote">... It seems all a reciprocating engine "really" does is produce a rotatating axis, if you will. It doesn't really produce ANY torque until a "lever" is attached to that axis and then we can measure the way the force acts on the lever producing what we call "torque" So without the "lever" there is no way to measure the hypothical torque that is in the spinning axis So the lever in our engine example would be an output shaft.</TD></TR></TABLE>That output shaft is the crankshaft itself.

The engine DOES produce torque. If it didn't, you could grab the crankshaft with your hand & stop it. Wait, don't try that... Torque is simply a twisting force. The transmission can't make torque or power. It can only convert it by the gear ratio.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">Machine B produces more torque than machine A. Piston engines make torque thru the pistons' downward force acting on the crank. The only way to increase torque is to increase the displacement of the engine, or increase the pressure inside the cylinders. Go back to my earlier post.

If you include the transmission within your concept of 'machine', then you can increase torque by lowering the gear ratio. So your description of A & B might actually be the SAME motor with different gearboxes. I've actually got 3/8" and 1/2" drills that look like the same motor with different gearboxes, so that's a good example.

That output shaft is the crankshaft itself.

The engine DOES produce torque. If it didn't, you could grab the crankshaft with your hand & stop it. Wait, don't try that... Torque is simply a twisting force. The transmission can't make torque or power. It can only convert it by the gear ratio.
</TD></TR></TABLE>
Thanks for the explaination. Now I understand
I think. LOL

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">The only way to increase torque is to increase the displacement of the engine, or increase the pressure inside the cylinders.</TD></TR></TABLE>
True. But have you seen those small engines that produce small amounts of horsepower but much more torque. How do they engineer the motor to produce more torque than horsepower?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nightryder21 &raquo;</TD></TR><TR><TD CLASS="quote">
True. But have you seen those small engines that produce small amounts of horsepower but much more torque. How do they engineer the motor to produce more torque than horsepower?</TD></TR></TABLE>


longer stroke, shitty r/s ratio. i know some ford engines use a r/s ratio of 1.45:1. VW engines have long strokes. itsa all in the angle and distance the rod travels in the engine. More severe angle more lowend, longer it travels the more torque as well.

 

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


longer stroke, shitty r/s ratio. i know some ford engines use a r/s ratio of 1.45:1. VW engines have long strokes. itsa all in the angle and distance the rod travels in the engine. More severe angle more lowend, longer it travels the more torque as well.</TD></TR></TABLE>
I was going to ask about this. Isn't the amount of "torque" an engine produces depenent on it's "squareness" as determinded by it's bore vs. stroke as posted above. Or as it's called it's "rod ratio"? Don't "undersquare" engines produce more torque than square or oversquare engines, all else being equal? IE. number of cylinders, ccd, induction method?

 

since horsepower is a function of torque and RPM, you can easy set any engine that can rev above 5252rpm (cross over point where HP is always greater than torque) to produce its peak torque past 5252 and thus peak HP is always greater than peak torque.
if you set it peak torque at a low rpm like 3000-4000 (like some trucks) so the torque value will usually be higher than the HP value.

all about how they want to set it.

I still think given the same displacement, 2 motors, one w/ a large bore and short stroke and one w/ a small bore and long stroke, all else the same the one with the longer stroke would produce more specific torque. While the one w/ the shorter stroke should be able to rev much higher, and possibly produce more peak HP since it can rev much higher.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by TypeSH &raquo;</TD></TR><TR><TD CLASS="quote">I still think given the same displacement, 2 motors, one w/ a large bore and short stroke and one w/ a small bore and long stroke, all else the same the one with the longer stroke would produce more specific torque. While the one w/ the shorter stroke should be able to rev much higher, and possibly produce more peak HP since it can rev much higher.</TD></TR></TABLE>


You are correct, that's why Formula 1 cars rev past 18K rpm and that's why tractors don't go over maybe 4K rpm, tractors need low speed torque so they can work properly and Formula 1 cars need to go fast.

Everything is a trade off, what do you want?

 

I think you can take ANY engine & make a cam for it that will either work good at low rpm or high rpm. But not both.

So since an oversquare engine is capable of spinning really high rpms without falling apart, that's why you usually see them with a high-revving cam. It's pointless to take an engine that can't spin 10,000rpm and design a cam that's optimized for 13,000.

I'm still gonna insist that torque comes from displacement and cylinder pressure. Nothing else. Whatever anybody does to increase torque is really increasing either displacement or pressure.

 

Edited for brilliance...

Another thought,

Displacement = Pi (Bore/2)^2*Stroke

We increase torque when we increase stroke because we are increase the radial distance from the axis at the crankshaft: the r in Torque = r X F. However, how exactly does increasing the bore (the diameter of the cylinder) increase torque. The radial distance of the rod attahced to the crank is not affected by the surface area of the piston. However, by increasing the surface area of the piston, we will increase the cylinder pressure (Force = int[Pressure*da,a] or Pressure = Force/area).

So to increase torque, you either increase piston force or increase stroke.


Modified by GSpeedR at 12:52 PM 8/21/2003

 

torque is the twisting motion of the engine. for general info a dyno measures torque and then uses a mathmatical equation to come up with an actual horsepower number. on smaller higher reving engines (two strokes specifically) flywheel weights can be added to crankshaft causing the engine to rpm slower because of the more rotational mass. this is just an example of how to use torque to "tune"

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">I think you can take ANY engine & make a cam for it that will either work good at low rpm or high rpm. But not both.</TD></TR></TABLE>

Correct, but you have to keep in mind (which you did) that the engine has certain limits that are determined by the bottom end and valvetrain.


<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">I'm still gonna insist that torque comes from displacement and cylinder pressure. Nothing else. Whatever anybody does to increase torque is really increasing either displacement or pressure.</TD></TR></TABLE>


Yeah, hondas can't expand displacement that much, but for cilinder pressure we can do a lot, that's why you have things like coatings, higher compression ratio pistons, longer rods, higher lift cams, timing control, high octane gas, etc....
They all increase the cilinder pressure hence they make torque and HP numbers a lil higher.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">I think you can take ANY engine & make a cam for it that will either work good at low rpm or high rpm. But not both.
</TD></TR></TABLE>

That is when Vtec comes into play, one cam profile for low rpm torque and another one that works best producing high rpm hp

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">... However, by increasing the surface area of the piston, we will increase the cylinder pressure (Pressure = int[Force*da,a] or Pressure = Force*area). </TD></TR></TABLE>You've got that backwards. Swap force with pressure...
FORCE = pressure * area

So if you increase bore, at the same stroke, same cylinder pressure; you increase the downwards force on the connecting rod. That increases torque, but you've increased displacement.

Increase stroke, at the same bore, like you said, increases torque. But it also increases displacement.

If you increase stroke but want to keep the displacement the same, you have to decrease bore. The original question (I think) had to do with increasing torque at the same displacement. Do the arithmetic - it works.

GZERO...
'Coatings' sounds interesting. What's that? Something to reduce friction in the engine? You can think of all this other stuff as increasing power, but friction 'uses up' some of that power before it even gets to the crank. All the other stuff you mention are ways to increase pressure.

jetpilot...
Yeah, VTEC is nice , but what that REALLY is, is like having 2 different camshafts to choose from. Lock it into the VTEC cams & tell me how it idles... Otherwise a slightly undersquare engine like Honda would not be a high-rev screamer. I mean, normally racing engines have to tolerate a really crappy idle & low-rev behavior in order to get all that power at the high end.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by TypeSH &raquo;</TD></TR><TR><TD CLASS="quote">since horsepower is a function of torque and RPM, you can easy set any engine that can rev above 5252rpm (cross over point where HP is always greater than torque) to produce its peak torque past 5252 and thus peak HP is always greater than peak torque.</TD></TR></TABLE>

HAHA yeah, the best way to make an engine produce more torque than horsepower is to detune the hell out of it and set the redline at like 3000rpm. Works for the domestics!!

 

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

Displacement = Pi (Bore/2)^2*Stroke

We increase torque when we increase stroke because we are increase the radial distance from the axis at the crankshaft: the r in Torque = r X F. However, how exactly does increasing the bore (the diameter of the cylinder) increase torque. The radial distance of the rod attahced to the crank is not affected by the surface area of the piston. However, by increasing the surface area of the piston, we will increase the cylinder pressure (Pressure = int[Force*da,a] or Pressure = Force*area).

So to increase torque, you either increase cylinder pressure or increase stroke.


Modified by GSpeedR at 8:05 PM 8/20/2003</TD></TR></TABLE>

Like JimBlake said, it's not pressure but force that's important. You could have 100000000psi in the cylinder, but if the bore is .0000000000001in it won't do anything for you. Increasing cylinder bore will make the same cylinder pressure work over a larger area, creating more force. Torque = force x lever arm distance.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">You've got that backwards. Swap force with pressure...
FORCE = pressure * area</TD></TR></TABLE>

Not one of my finer moments...

So we increase the bore, we increase the surface area, and we increase the force, which increases the torque. We didn't touch cylinder pressure in doing so (Fack!).

So we either increase the stroke or increase the piston force.

So most Honda engines will increase the bore (while leaving the stroke the same) when increasing displacement. Is this to keep piston velocities down, which would cause huge stress at high rpm?


Modified by GSpeedR at 1:13 PM 8/21/2003

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">So we increase the bore, we increase the surface area, and we increase the force, which increases the torque. We didn't touch cylinder pressure in doing so (Fack!).

So we either increase the stroke or increase the piston force. </TD></TR></TABLE>Increase the bore, which increases torque. Yes, but you've increased displacement. Same thing with stroke. The original question sounded (to me ) a little more theoretical. If you wanted to increase the bore without changing the displacement, you have to decrease the stroke to compensate. That puts the torque pretty much back where you started.

In real life nobody who bores out their engine is going to shorten their stroke, because you don't HAVE to keep displacement the same. Well, maybe if you're under the rules of some sanctioning body...

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">So most Honda engines will increase the bore (while leaving the stroke the same) when increasing displacement. Is this to keep piston velocities down, which would cause huge stress at high rpm?</TD></TR></TABLE>I don't know how 'most' Hondas do it. When you compare B16 vs. B18 vs. B20? When you compare F23 vs. some other displacement?

Generally oversquare engines can rev higher because of piston velocities. But bigger bore means heavier pistons, so as far as stress goes, the answer is 'it depends'...

When you're talking about Honda designing a new displacement under an existing engine architecture, it's one thing. When you're talking about a car OWNER deciding to bore or stroke HIS engine, that's a different story. I'm beginning to wonder if this is really what the original question was all about...

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">I'm beginning to wonder if this is really what the original question was all about...
</TD></TR></TABLE>
IMHO, I thought the original question was naieve and was asking
why aren't ALL Honda engines capable of producing MORE torque
than they do in reality. But I'm probablly wrong. LOL
On re-reading the original post it seems the questions is where does "torgue" come from and how do honda owners get "more". LOL
As if torgue is something magical you can just buy and bolt to an engine. But as I said above I could be wrong. LOL

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">Increase the bore, which increases torque. Yes, but you've increased displacement. Same thing with stroke. The original question sounded (to me ) a little more theoretical.</TD></TR></TABLE>

I don't remember restrictions on "displacement", so I just went for the simplest scenario (I'm a physics person rather than engineer...for now).

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote"> life nobody who bores out their engine is going to shorten their stroke, because you don't HAVE to keep displacement the same. Well, maybe if you're under the rules of some sanctioning body...</TD></TR></TABLE>

There are bodies based upon displacement (PCA under 2.5L), but I will definitely agree, that nobody is going to shorten the stroke to compensate for a larger bore (and especially not vice versa).

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Generally oversquare engines can rev higher because of piston velocities. But bigger bore means heavier pistons, so as far as stress goes, the answer is 'it depends'...</TD></TR></TABLE>

That was the answer I was looking for.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">GZERO...
'Coatings' sounds interesting. What's that? Something to reduce friction in the engine? You can think of all this other stuff as increasing power, but friction 'uses up' some of that power before it even gets to the crank. All the other stuff you mention are ways to increase pressure.
</TD></TR></TABLE>

Yes, coatings (IMHO, maybe some other people would not agree with me) can lower friction and also reflect heat into the combustion chamber, but there are a lots of coatings to choose from. From thermal coatings to friction coatings, i think they are the best and they are becoming more popular with time. we'll see

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote"> So we increase the bore, we increase the surface area, and we increase the force, which increases the torque. We didn't touch cylinder pressure in doing so (Fack!). </TD></TR></TABLE>

well, not true, when you increase the bore you increase the compression, hence you increased the force.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">So most Honda engines will increase the bore (while leaving the stroke the same) when increasing displacement. Is this to keep piston velocities down, which would cause huge stress at high rpm?</TD></TR></TABLE>


Yes, messing with the bore wouldn't affect rod to stroke ratio, hence they won't mess with piston velocity.

 

I thought the Original question was what components or modificationcs can be made to produce MORE torque.

Some one said Increase Displacement and then this sparked the discussion on WHY displacement will effect torque.

Someone hit it on the head pretty directly saying that... Increasing the Rod stroke adds more leverage and thus increasing the torque, but then why do you increase TORQUE when you over bore. And it was explained that there is more area to apply the downward force.
I know that stiffer valve springs increase the torque but I don't know how exactly to explain it.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GZERO &raquo;</TD></TR><TR><TD CLASS="quote">well, not true, when you increase the bore you increase the compression, hence you increased the force.</TD></TR></TABLE>

Assuming that the pressure stays the same, we will increase the force simply by increasing the surface area. However, since the compression doesn't stay the same, but actually increases a little (I had to find the C-ratio formula to check that), force will increase more, though (correct me if I'm wrong here) not a whole lot. The fact that the piston area has a larger surface area multiplies the force to a greater extent than the fact that the compression ratio has changed.

Force = Pressure*(bore/2) ...very simple forumla...

C-ratio = Pi*(bore/2)^2*((stroke + [compressed height])/[compressed height])

The term [compressed height] includes head gasket thickness, added deck height combustion chamber, and piston top volume.

So a 1 mm increase in bore yeilds a 0.5 increase in force (units?).

So a 1 mm increase in bore yeilds a 0.25 increase in compression ratio.

I haven't gone through to attempt to figure out how the increase in compression is related to the force at the piston (I'm sure it depends on a LOT), but large increases in bore greatly increase C-ratio (ratio increases with the square of the cylinder radius rather than linearly as in the "force"). To save me the trouble, has anyone figured this out? Or am I missing something big?


Modified by GSpeedR at 8:33 PM 8/22/2003

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">C-ratio = Pi*(bore/2)^2*((stroke + [compressed volume])/[compressed volume])</TD></TR></TABLE>Gotta re-arrange the parentheses there...

CR = (Vc + disp) / Vc
where:
disp = displacement = pi*(bore/2)^2 * stroke
Vc = your 'compressed volume' which includes combustion chamber volume corrected by headgasket & piston features.

Boring an engine, keeping the same head, can increase CR. But re-designing an engine for larger bore, you presumably enlarge the chamber in the head, because then you can fit bigger valves.

Just another example of when you change anything you probably change a couple other things you didn't think about. Increasing compression ratio is one way to increase cylinder pressure.

While the original question was probably about 'how can I get more torque', I was trying to clarify some stuff that sounded like smoke & mirrors. There's a lot of technical jargon being used by people without a physics background. I guess instead of saying 'that's dumb' I wanted to try & explain some of it.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JimBlake &raquo;</TD></TR><TR><TD CLASS="quote">I'm beginning to wonder if this is really what the original question was all about...</TD></TR></TABLE>

The idea of the topic was to get all useful information about torque. I know a lot of people that have questioned what and where torque comes from. Now anybody wondering about the phenomenom(I know spelled it wrong, I think.) that is called torque just has to look at this post and get preatty much a lot of useful info. A lot of unasked questions were answered here. Any questions about torque now should hopefully be answered with a link to this post. Just really look back at the post and see how much info was covered. I gave you guys a bone and you guys ran with it.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Spoond TEG &raquo;</TD></TR><TR><TD CLASS="quote"> I know that stiffer valve springs increase the torque but I don't know how exactly to explain it.</TD></TR></TABLE>

Stiffer valve springs won't increase torque. In fact, they might actually decrease it due to increased friction.

They do however allow you to run bigger cams, among other things, and those might increase torque.