Stroke, Bore, Torque & Power

[skippybangkok]

My Nephew is studying automotive ( diploma level ), and we got into a discussion about the above topic.

In theory - if you increase the crank arm length, the "arm" becomes longer, giving it more torque - BUT, it would increase the displacement, and as such, if you want to compare same displacement (cc's ) the Bore would have to reduce to maintain the original displaement. A reduction in Bore, would reduce force on piston - Summary - if u run some numbers in Excel, the Torque will remain the same because a longer arm ( more torque ) means smaller piston ( less torque )........... and the numbers come out identical.

The only thought i have is the E (Engery ) = Torque x tau ( angula speed of crank ). I suspect the crank would go slower for a larger arm, reducing angualr speed, but increasing torque at a fixed E ( fixed displacement = fixed amount of air / fuel burnt ).

Anyone can explain this?

[dave_boo]

My Nephew is studying automotive ( diploma level ), and we got into a discussion about the above topic.

In theory - if you increase the crank arm length, the "arm" becomes longer, giving it more torque - BUT, it would increase the displacement, and as such, if you want to compare same displacement (cc's ) the Bore would have to reduce to maintain the original displaement. A reduction in Bore, would reduce force on piston - Summary - if u run some numbers in Excel, the Torque will remain the same because a longer arm ( more torque ) means smaller piston ( less torque )........... and the numbers come out identical.

The only thought i have is the E (Engery ) = Torque x tau ( angula speed of crank ). I suspect the crank would go slower for a larger arm, reducing angualr speed, but increasing torque at a fixed E ( fixed displacement = fixed amount of air / fuel burnt ).

Anyone can explain this?

I'll try.

First you're over considering the effect that the stroke will have on the displacement. Volume for a cylinder is pi*r²*height. Let's consider a CBR 150's bore/stroke/volume. Bore is 63.5 and stroke is 47.2 mm. Overall displacement is 149.5cc. If you go and add 5 mm to the stroke (and want to stay at the same displacement), your bore is only going to shrink some 3.3 mm. Getting a bigger engine and increasing the stroke even more will result in a propotionate ratio (i.e., you only need to reduce the bore some 60%).

Added on top of that is the weight. Connecting rods are MUCH heavier than flyweight pistons. The addition of an extra 5 mm to the conrod's length way over compensates for the slight loss in weight the 3.3 mm that was removed from the bore (and therefore the piston) to achieve displacement parity. This weight adds torque.

It looks like you may be confused about Boyle's Law also. Increasing the distance that the piston moves will, assuming similar deck heights/piston heads/domes/etc., will allow a more gradual increase in pressure at it arises to TDC and another decrease in pressure as it drops back down. These finer grained pressure drops allow a 'better' cam design that allows more air to be drawn in and forced out. More air=more power.

You were on the right track about angular velocity though; longer stroke means that, at the same RPM for a short stroke, the crank's outer edge, where the piston is attached to, is going faster. The added mass on the crank that is needed for the extra stroke also helps (F=M*A).

[skippybangkok]

Thanks.......... a few idea's / brainstorming

Added on top of that is the weight. Connecting rods are MUCH heavier than flyweight pistons. The addition of an extra 5 mm to the conrod's length way over compensates for the slight loss in weight the 3.3 mm that was removed from the bore (and therefore the piston) to achieve displacement parity. This weight adds torque.

Good point.....but this is inertia not torque. Torque = Force x Arm ( no mention of mass in the formula ) . As per the original post, and as per your example, a increase in arm would require reduction in bore ( T = Fx a ), and the torque would not change ( one up, one down, end result no change ). Good point though that the weight would be different - thus acceleration of an engine with small stroke would better than large stroke engine. But once up to steady state ( constant RPM ) this would not affect torque (???)

It looks like you may be confused about Boyle's Law also. Increasing the distance that the piston moves will, assuming similar deck heights/piston heads/domes/etc., will allow a more gradual increase in pressure at it arises to TDC and another decrease in pressure as it drops back down. These finer grained pressure drops allow a 'better' cam design that allows more air to be drawn in and forced out. More air=more power.

Dont believe Boyles Law is correct (isothermic compression) .... it would be an Adiabatic Compression ( T2 = T1·Rc(k-1)/k ). Beside the point really.....agreed a nice long stroke would be more effective in sucking in and dissapating air/fuel mixture.

Thanks for your help..... the piston weight v.s. crank weight would help in the acceletation of an engine, but i dont feel that it would affect the Torque. Still pondering why............

........then again........ there will be constant acceleration and deceleration as the cylinder goes from detonation to the point its ready to be fired again...........

Confuddled.

[dave_boo]

Thanks.......... a few idea's / brainstorming

Good point.....but this is inertia not torque. Torque = Force x Arm ( no mention of mass in the formula ) . As per the original post, and as per your example, a increase in arm would require reduction in bore ( T = Fx a ), and the torque would not change ( one up, one down, end result no change ). Good point though that the weight would be different - thus acceleration of an engine with small stroke would better than large stroke engine. But once up to steady state ( constant RPM ) this would not affect torque (???)

I beg to differ; mass is implicitly mentioned by using the term "Force". Force=Mass*Acceleration is a standardised formula. Also, it appears that you're confused about the torque formula. The formula that you're looking at is better suited for imagining torque on a lever than on a piston. Granted it is ok to think about a piston using it, but you're assuming (from the way I read your post) "Arm" in this case refers to the conrod. The conrod is actually not the "arm", but rather the length from the centre of the crank to the centre of the lobe the piston attached to is the "arm". Think about it; that's the part that moves, not the conrod (who's movement is dependent on that "arm"). There will be some multiplication of the force since you're expecting the piston to travel up and down, and this multiplication will be even greater for a longer conrod. So, even totally disregarding my point about the weight and all that, plugging in the greater stroke (notice there's no mention of volume in your equation) will automatically result in a greater torque. This is super simplified in ignoring mechanical losses. And yes, a smaller stroke engine will spin up faster and be able to sustain higher RPMs. However, at the same RPM as the longer stroke engine the torque will be down. Hitting those higher RPMs will allow it to reach higher HP numbers and that can be benifical in certain circumstances; however the design of components has to be different to eek out that power (resulting in a loss of power down low) because an internal combustion engine is NOT capable of nice linear power gains across the RPM range.

Dont believe Boyles Law is correct (isothermic compression) .... it would be an Adiabatic Compression ( T2 = T1·Rc(k-1)/k ). Beside the point really.....agreed a nice long stroke would be more effective in sucking in and dissapating air/fuel mixture.

Boyle's Law works well if the engines are running at the same speed; they should be really close to the same temperature and the intake charge will be largely the same to safely ignore the adiabatic compression. Granted the greater rotating mass may generate more heat, but that may be offset by the slightly longer time it takes the piston to reach each end of the stroke and the increased air charge cooling. But that's a bit of a guess on my part. It's also my understanding that the constant volume adiabatic process is used to formulate temperature/pressure in petrol powered engines; since you insisted on comparing two similar displacements the temperature/pressure should be the same, once again assuming similar decks/piston heads/domes. The Boyle's law comes into play whilst the piston is in motion; not at BDC/TDC.

Thanks for your help..... the piston weight v.s. crank weight would help in the acceletation of an engine, but i dont feel that it would affect the Torque. Still pondering why............

[dave_boo]

Ok, a quick idea for you to understand. Get a piece of string a meter long and tie a largish can to it. Spin it around.

Now find a can that's 1/2 as large. Tie a string 1,8 meters to it. Spin it around.

Tell us your findings.

[dave_boo]

Another couple of things I left out in my attempts to keep it abbreviated; volumetric, thermal, and mechanical efficiency.

Volumetric efficency can be improved by having a more narrow bore in your example. The decrease in cavitation and eddies as the piston drops on the intake stroke can be detrimental to efficency. A more narrow bore doesn't have as much horizontal space to expand into and should draw the air more smoothly. Air drawn in more smoothly means more air in and more power.

Thermal efficiency can be improved due to a smaller diameter piston used in a longer stroke engine. With less surface area there's less heat picked up each combustion (and retained) and thus the intake charge remains cooler.

Mechanical efficency on a smaller diameter piston in a longer stroke is pretty self explanatory. Going back to my CBR 150 piston, the standard piston circumference is 199.5 mm; increase your stroke 5 mm and decrease your bore to 60.2 mm nets you a circumference of 189.1 mm. An extra 10.4 mm of drag inducing rings is quite a bit. This is of course ignoring power loss due to other factors.....but every little bit helps.

**edit**

Didn't see this part before, but it's worth a comment:

........then again........ there will be constant acceleration and deceleration as the cylinder goes from detonation to the point its ready to be fired again...........

This ties into to the mechanical efficiency I covered earlier. Let's say that for the sake of argument the crank has to move 1/8 of a rotation before it actually moves the piston from TDC. Now, on the shorter stroke, the circumference of the circle transcribed by the crank is 296.6 mm. The 5 mm longer stroke crank has a circumference of 328 mm. 1/8 of a rotation means the short stroke travels 37.1 mm before there's piston movement and the long stroke travels 41 mm. This adds momentum, especially useful at the begining of the exhaust cycle. Furthermore it should 'ease' the piston out of its inertia better than the quicker short stroke, lessing the force per time period to move it. I hate to bring yet another physical experiment in; but make a fist and circle your hand in front of your body in a small circle every second. Repeat with a larger circle. Which one put less pressure on your elbow? The larger one should put less pressure on it leading you to the conclusion that a large stroke engine will also require less power to do the same.

[BSJ]

Can the OP re-phrase the question, the original post is leading to confusion amongst the answers!

[dave_boo]

Can the OP re-phrase the question, the original post is leading to confusion amongst the answers!

Perhaps I'm misunderstanding, but let me try to explain what he's asking.

On the assumption that you're American, why does a long stroke short bore engine (think a Chevy small block 5,4 L-350cc) and a Ferrari short stroke (5,4L) not make the same amount of torque?

I've tried to answer that; but let me create a confusion (hopefully!) free post that condenses my answers down.

A.) There's mechanical advantages to a longer stroke resulting from the decreased friction from piston rings, leverage upon movement of the piston (and 'easing' it out of where it was), interia of the added mass from the longer conrods and crankshaft, etc.

B.) There's volumetric efficiency to a longer stroke resulting from the decreased turbulence on the intake and exhaust stroke.

C.) There's thermal efficiency to a longer stroke resulting from a decreased area that the air charge can pick up heat from.

D.) An internal combustion engine is not linear in power increases. Playing with various parts can approximate this, but you can not have both low down torque and high end HP.

[hansnl]

Can the OP re-phrase the question, the original post is leading to confusion amongst the answers!

Perhaps I'm misunderstanding, but let me try to explain what he's asking.

On the assumption that you're American, why does a long stroke short bore engine (think a Chevy small block 5,4 L-350cc) and a Ferrari short stroke (5,4L) not make the same amount of torque?

I've tried to answer that; but let me create a confusion (hopefully!) free post that condenses my answers down.

A.) There's mechanical advantages to a longer stroke resulting from the decreased friction from piston rings, leverage upon movement of the piston (and 'easing' it out of where it was), interia of the added mass from the longer conrods and crankshaft, etc.

B.) There's volumetric efficiency to a longer stroke resulting from the decreased turbulence on the intake and exhaust stroke.

C.) There's thermal efficiency to a longer stroke resulting from a decreased area that the air charge can pick up heat from.

D.) An internal combustion engine is not linear in power increases. Playing with various parts can approximate this, but you can not have both low down torque and high end HP.

For me it certainly is a very long time ago.

However, I seem to remember that long stroke engines can run with a lower RPM to give the same amount of torque.

I do also remember that the (in)famous Citroen DS had a long stroke engine to make the engine quiter, lower rpm.

So I guess you are right.

And forgive me if I am wrong.

[skippybangkok]

Can the OP re-phrase the question, the original post is leading to confusion amongst the answers!

Perhaps I'm misunderstanding, but let me try to explain what he's asking.

On the assumption that you're American, why does a long stroke short bore engine (think a Chevy small block 5,4 L-350cc) and a Ferrari short stroke (5,4L) not make the same amount of torque?

Yep - thats the question.

[skippybangkok]

So, even totally disregarding my point about the weight and all that, plugging in the greater stroke (notice there's no mention of volume in your equation) will automatically result in a greater torque.

Will go bit by bit.............. As per the above, if same bore - agreed (but then the diplacement would be more. - to maintain same displacement, you would need to reduce the bore ( reducing force ) - I have modelled it in a spreadsheet. Arm (stroke ) goes up, Bore down - both cancel each other out

I am trying to (not correctly) ignore combustion efficiencies for the time being and resistance

The only high level conclusion i have so far is from Torque Write Up which puts it in a simple way

1. E (Energy) = Torque x Angular Rotation speed ( rpm in rads ). They simply stated the longer arm is geared differently to reduce the RPM, and thus following the formula = increase torque ( wrongly assuming the Air/Fuel mixture will be the same for the same displacement )

2. Optimum torque, because of combustion characteristics, will be at about 3000 RPm give or take.

I guess what really caught me by surpise was that the theoretical torque was the same for a long and short stroke engine at the same displacement.

[Lopburi99]

Man, it's been years and years since I've studied these subjects. Good posts for my gray matter as I try to remember, understand and follow your logic and the theory. Thanks guys.

Just a related thought though. I once had a V12 Jaguar 5.3 liter. Talk about smoothness and low end torque! Quiet and powerful - awesome!

[skippybangkok]

Yeh...bit of fun...not bothered who is wrong right...... just wanna crack this nut........... ( cant be shown up by my nephew ! - so must be right )

and extract from Myth #5 in my above URL:-

The torque applied to the crank depends on the force as well as the crank throw distance. The force that the gas exerts on the piston face is proportional to both the gas pressure and the area of the piston face. As the stroke is make longer, for a given displacement, the piston face area is made proportionally smaller, and the force exerted on the piston face is made proportionally smaller. Thus, the effect of increasing the stroke length is cancelled by the coupled effect of reducing the piston face area. Here, we see that even the maximum instantaneous torque applied to the crank will be independent of the stroke if the stroke variation is subject to a constraint on the displacement.

Instead of the long stroke causing the engine to produce more torque at low engine speed, the truth of the matter is that the long stroke and its associated greater piston speed and piston acceleration limit the engine speed. As such, it only makes sense that the design characteristics such as the valve lift and duration be optimized for greatest volumetric efficiency at the lower engine speeds where that engine will always be operated. Because of that specific optimization, you would expect that such an engine should be capable of producing more torque at those low engine speeds than an engine that is not similarly optimized for low engine speed.

However, power determines acceleration, and power depends on the engine speed, so an engine of this sort is inherently incapable of achieving the same power or acceleration as compared to a high rpm engine of similar displacement, at any road speed. Even "off the line", if a short stroke, high rpm engine is mated to a gear box with 1st gear set adequately low, the short stroke, high rpm engine will out-accelerate the long stroke, low rpm engine, anytime, anywhere

So....it seems Dave_boo's discussion on volumetric efficiency and better aspiration coupled with the lower RPM is the main cause of torque - not the lenght of the arm. ( crank center / Rod connection to crank offset )

[dave_boo]

Will go bit by bit.............. As per the above, if same bore - agreed (but then the diplacement would be more. - to maintain same displacement, you would need to reduce the bore ( reducing force ) - I have modelled it in a spreadsheet. Arm (stroke ) goes up, Bore down - both cancel each other out

I am trying to (not correctly) ignore combustion efficiencies for the time being and resistance

The only high level conclusion i have so far is from Torque Write Up which puts it in a simple way

1. E (Energy) = Torque x Angular Rotation speed ( rpm in rads ). They simply stated the longer arm is geared differently to reduce the RPM, and thus following the formula = increase torque ( wrongly assuming the Air/Fuel mixture will be the same for the same displacement )

2. Optimum torque, because of combustion characteristics, will be at about 3000 RPm give or take.

I guess what really caught me by surpise was that the theoretical torque was the same for a long and short stroke engine at the same displacement.

If you were to choke off the intake (and I know that you're attempting to ignore it), than yes, your line of thought is correct-a shorter stroke will yield the same torque. However, quoting from your article (mainly because it re-enforces what I said earlier):

To the extent that a long stroke engine happens to produce more engine torque at low engine speed as compared to a short stroke engine, the reason for this is at best indirectly related to the length of the stroke. Rather, it can only be due to a difference in volumetric efficiency at that lower engine speed, i.e., design characteristics such as valve lift/duration and the rate of volumetric expansion of the combustion chamber on the intake stroke.

Your point 1.) is merely a way of illustrating that a higher torque engine will produce power at a lower RPM than a lower torque engine. Following this line of thought means that eventually the lower torque engine (which can rev higher) will eventually meet, and quite likely exceed, the peak power that the higher torque engine produces; just at a higher RPM.

Yeh...bit of fun...not bothered who is wrong right...... just wanna crack this nut........... ( cant be shown up by my nephew ! - so must be right )

and extract from Myth #5 in my above URL:-

The torque applied to the crank depends on the force as well as the crank throw distance. The force that the gas exerts on the piston face is proportional to both the gas pressure and the area of the piston face. As the stroke is make longer, for a given displacement, the piston face area is made proportionally smaller, and the force exerted on the piston face is made proportionally smaller. Thus, the effect of increasing the stroke length is cancelled by the coupled effect of reducing the piston face area. Here, we see that even the maximum instantaneous torque applied to the crank will be independent of the stroke if the stroke variation is subject to a constraint on the displacement.

which was my finding ( to my suprise) with the excel formula i made.

This article was written with an extreme low torque/high HP bias. Going back to the CBR piston, the area lost in maintaining the displacement but increasing the stroke by 5mm (and decreasing the bore by 3,3 mm) was all of 9,9% (3 166,9 mm² versus 2 855,8 mm²). Note however, you gained 10,6% more stroke.

Please note that they talked about instateneous torque. Further up in the article it described that and average torque.

The instantaneous torque varies considerably throughout that rotation, even within the ½ crank rotation corresponding to the power stroke. The quantity of work done during one complete rotation of the crank is fully determined by the integral of the instantaneous torque over that complete rotation. The average torque over the rotation is the integral of the instantaneous torque divided by the angular distance, so it follows that the average torque over the crank rotation effectively determines the quantity of work done over that crank rotation.

This is, IMHO, an over-simplification that ignores mass (amongst other things). If all the bits and bobs were truely weightless, than yes, you could use that line of thought. However, the mere fact that the instateneous torque has to overcome interia of the heavier crank and conrods and that force is than carried through 3 more strokes, merely by the interia of the aforementioned parts, clearly illustrates my point. Lighter parts require less force to change their interia and retain less force. Going heavier with a longer stroke retains that interia.

If you want to show your nephew up, remind him you helped Sir Issac Newton come up with his laws of motion (that apple didn't fall by itself!). F=M*A. Torque is a force. Acceleration is described, in this case, by RPM. Mass can be deduced. This is the absolutely simplest way of demonstrating it; removes as many variables as possible. If you want to get more detailed than that, you're going to have to go with my post here and expound on it further with the maths that describe the increased interia from both the longer conrods and the greater mass of the crank (and quote possibly a heavier harmonic balancer), force upon the decreased surface area, reduced friction from the rings, reduced resistance to interia from a lighter piston, reduced resistance to interia from smaller piston pins, increased intake and exhaust times, efficiency of the expansion of air charge, decrease in time of flame edge propogation due to less surface area, etc.

[skippybangkok]

Cheers....he probably believes i used to know Isac Newton in person.........

[PeaceBlondie]

This evokes a question from my past: the only Ferrari I've ever touched was at age 15, a Monza 750 racer. http://en.wikipedia.org/wiki/Ferrari_Monza That model won many races against 6-cylinder Mercedes and Jaguars. and V-12 Ferraris. I'm no engineer, but always understood that larger cylinders (like a 750x4) had better torque than smaller cylinders (250x12). With exceptions. Of course, short strokes allow higher rpm.

[Lopburi99]

This evokes a question from my past: the only Ferrari I've ever touched was at age 15, a Monza 750 racer. http://en.wikipedia.org/wiki/Ferrari_Monza That model won many races against 6-cylinder Mercedes and Jaguars. and V-12 Ferraris. I'm no engineer, but always understood that larger cylinders (like a 750x4) had better torque than smaller cylinders (250x12). With exceptions. Of course, short strokes allow higher rpm.

It will be interesting to hear what skippy and dave have to say about that PB. Also, I would think the number of cylinders, and therefore the number of firings per crankshaft revolution, comes into the overall engine torque equation somewhere.

[skippybangkok]

This evokes a question from my past: the only Ferrari I've ever touched was at age 15, a Monza 750 racer. http://en.wikipedia.org/wiki/Ferrari_Monza That model won many races against 6-cylinder Mercedes and Jaguars. and V-12 Ferraris. I'm no engineer, but always understood that larger cylinders (like a 750x4) had better torque than smaller cylinders (250x12). With exceptions. Of course, short strokes allow higher rpm.

It will be interesting to hear what skippy and dave have to say about that PB. Also, I would think the number of cylinders, and therefore the number of firings per crankshaft revolution, comes into the overall engine torque equation somewhere.

The question never about adding or subtracting cylinders.....rather the impact of an under of over square engine with the same displacement. At a very basic calculation, both would have in theory the same torque......... so it was to find out why its not the case in reality. PB's statement is emperical knowledge ( based on experience), was trying to dig deeper as to why this was. Cheers

[dave_boo]

The question never about adding or subtracting cylinders.....rather the impact of an under of over square engine with the same displacement. At a very basic calculation, both would have in theory the same torque......... so it was to find out why its not the case in reality. PB's statement is emperical knowledge ( based on experience), was trying to dig deeper as to why this was. Cheers

I was looking more deeply into the article you linked to. It has some serious flaws to it, but there's something that jumps out immediately that backs up my position. A few forumlae are listed, to whit

Torque is measured by taking the right-angle distance from the line of force to the axis of rotation, and multiplying the force by that distance. Torque = Moment of Inertia x Angular Acceleration

Work = Torque x Angular Distance

Power = Torque x Angular Velocity

If we first define torque by determining that the moment of inertia will be greater due to the greater mass of the longer rods and heavier crank, and we concede that the angular acceleration remains the same due to being the same RPM, than it's rather simple to understand that the formula works out to the favour of the longer stroke engine. Solving for work and power further re-enforce this since the angular distance transcribed by a longer throw crank and the increased velocity incurred by being further from the centre of the crank are both larger on the longer stroke engine multipling the spread between the two engine types.

[klikster]

This evokes a question from my past: the only Ferrari I've ever touched was at age 15, a Monza 750 racer. http://en.wikipedia.org/wiki/Ferrari_Monza That model won many races against 6-cylinder Mercedes and Jaguars. and V-12 Ferraris. I'm no engineer, but always understood that larger cylinders (like a 750x4) had better torque than smaller cylinders (250x12). With exceptions. Of course, short strokes allow higher rpm.

It will be interesting to hear what skippy and dave have to say about that PB. Also, I would think the number of cylinders, and therefore the number of firings per crankshaft revolution, comes into the overall engine torque equation somewhere.

I think that torque at what RPM is much of the issue.

[BSJ]

Post #13's red text gives a good explanation. We seem to be now on the road to mechanical harmony!

My preference is for short stroke engines of 500cc or less per cylinder with good breathing heads and racy cams.

Those long stroke engines are for plodding along...visualize old farts wearing floppy hats driving down to the bowling club.