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Killrob

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Post Fri Apr 22, 2011 6:14 am

Bore & Stroke Discussion

This post is going to try to discuss how to translate between bore&stroke and power&torque.

I have thought about this for a while and put quite some effort into this post. So here we go. First,
let us try to establish a footing for the terms torque and power. Because there are so many factors to
consider, I will focus just on stroke and bore and nothing else, keep that in mind while reading the following.

Also, if you are not familiar with engines, I recommend taking a look at reference [0] which has a very nice
illustration of the basic working principle in the form an animated figure.


Power

Power is generated by adiabatically [1] compressing a fuel/air mixture, combusting it and then letting
it adiabatically (isentropically? [2]) expand. The power output is generated by the difference between
a) the energy required for the compression, and b) the energy generated by the heated expanding gas in the
cylinder during/after the combustion process [3] during the so-called power stroke. How much energy
can be extracted from the expanding gas depends on how much the gas is allowed to expand. The first steps
of the expansion release more energy than the final steps of the expansion (or at least that seems logical to
me... but don't quote me on that ;) ).


Torque

Torque is produced by the turning of the crank... or more precisely: the force pushing down the piston
acts on the crank, which has an offset to the axis formed by the connection rods, and thus gives rise
to torque. The torque generated at any given moment (roughly) is the force on the piston times the
orthogonal offset of the crank to the axis.

M = F * r
Edit: this is for the orthogonal case only, else it's the orthogonal projection of the force vector which
is considered, i.e. M = r x F as Deus ex Machina pointed out correctly.

Easier than it sounds, the maximum offset from the connection rod axis is given by stroke/2. The
maximum torque is generated about half-way through the expansion phase as the crank becomes
orthogonal to the axis.

torque maximum = force on piston * stroke/2

This will not be maximum effective torque... which will be closer to the maximum torque divided by
the square-root of 2.

[0] http://en.wikipedia.org/wiki/4-stroke_engine
[1] http://en.wikipedia.org/wiki/Adiabatic_process
[2] http://en.wikipedia.org/wiki/Isentropic_process
[3] http://en.wikipedia.org/wiki/Four-stroke_engine


The following two simple examples will help us understand the dependencies and get them right.


Example a) bore increase

We have one cylinder with inner radius r, i.e. a bore of 2r. The stroke is s. If we now just increase
r (bore/2) to R (BORE/2) and leave all other parameters the same (while scaling the amount of fuel
injected), how changes the power output and torque of the engine?

We established that the power is created by the expanding gas. The difference in the amount of
gas is given by the difference in engine displacement. The engine displacement is given by:

pi * r^2 * s,

and the new displacement is

pi * R^2 * s.

As everything else is the same, except for the amount of fuel (as stated before), we have a linear
dependence between the force on the piston and the engine displacement. Thus, the force on the
piston of the engine increases (to first order) by a factor R^2 / r^2. The torque of the engine
depends on the force on the piston if the stroke is the same in both cases. Thus the engine torque
also increases by a factor R^2 / r^2. This scaling factor can be rewritten as

R^2 / r^2 = (BORE/2)^2 / (bore/2)^2 = BORE^2 / bore^2

Summary of example a)
Increasing the bore and just the bore, both power and torque scale approximately linearly with the
engine displacement.


Example b) stroke increase

We have one cylinder with inner radius r, or a bore of 2r. The stroke is s and is increased to S.
All other parameters (also the amount of fuel!) is left the same, how changes the power output and
torque of the engine?

The initial volume from where the gas expansion starts is the same as before. The force on the
piston is the same as before as well. What is different is the lever that produces the torque and the
amount the gas is allowed to expand before the power stroke ends. The difference in lever length
yields an increase in maximum torque from

M = F * s/2 to M = F * S/2

which is an increase by a factor of S/s. Simply a linear increase.

The extended stroke is more tricky. It makes the engine more efficient as more energy is extracted
from the expanding gas in total. Compared to the engine with stroke s, a force acts on the piston for
the original path + the additional path. The dependence here is not linear if the initial expansion is worth
more than the later stages of the expansion - this would mean it's "less than linear", and probably depends
on the gamma factor mentioned in references [1] and [2]. The actual dependence can be extracted
when knowing and understanding these references... (which I admit that I currently don't, even though
I should) and having some values to put in from known engines. Without this, it's just guessing... so let's
say it is 0.7, which would be "less than linear", which would be 1.0, but more than a square root
dependence, which is 0.5.

By making this wild guess of 0.7, the increase in power output would be a factor of

(S/s)^0.7

For example, by increasing the stroke from 50mm to 75mm, the power would increase by roughly 33%,
instead of 50% which would be the case for a linear dependence.

Also note that I did not speak of compression, which is important in this case because keeping
everything constant while increasing the stroke actually increases the compression value.

Summary of example b)
Increasing the stroke and just the stroke increases the torque of an engine linearly with the stroke,
while it increases its power output less than linearly.

Comment: this dependency can be determined by our engine experts as well, by just comparing
engines with equal bore but different stroke (although there are many more parameters...)


So much for that... the discussion is on you now ;)
Cheers!
Last edited by Killrob on Fri Apr 22, 2011 8:58 pm, edited 4 times in total.
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Deus ex Machina

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Post Fri Apr 22, 2011 7:15 am

Re: Bore & Stroke Discussion

The first steps of the expansion release more energy than the final steps of the expansion


that assumes that the explosion starts at the same time everywhere in the fuel/air mixture. it starts at the position of the spark plug, which normally is somewhere in the top (between the intake/exhaust valves); so the initial energy isnt the highest, the one when the whole mixture went boom is. (even if its just a few milliseconds afterwards).
but u r right if u meant the stress on the piston is the highest during this explosion phase.



also all those calculations are theoretical, since u have no friction in the formulas.

just when is the increase in friction higher, by increasing the bore (more circumfencial (sp?) area of the piston and also the cylinder) or the stroke (same area for the piston, but it rubs against more height (=area) of the cylinder)?



as a side note: M = F * r only works if u dont have vectors. with vectors u gotta use M = r x F, since a x b = -(b x a)
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Killrob

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Post Fri Apr 22, 2011 7:48 am

Re: Bore & Stroke Discussion

Deus ex Machina wrote:
The first steps of the expansion release more energy than the final steps of the expansion


that assumes that the explosion starts at the same time everywhere in the fuel/air mixture. it starts at the position of the spark plug, which normally is somewhere in the top (between the intake/exhaust valves); so the initial energy isnt the highest, the one when the whole mixture went boom is. (even if its just a few milliseconds afterwards).
but u r right if u meant the stress on the piston is the highest during this explosion phase.


Yes, you are right in that I assume that the energy is created at time zero, which is not entirely correct as you point out. Yet I think it is a valid approximation. But I think you misunderstand my true meaning: if you take a look at http://en.wikipedia.org/wiki/Thermodynamic_process you see a diagram on the left side. In this diagram I think I remember that more energy is released along the first part of the "1" trajectory than in its later part.


also all those calculations are theoretical, since u have no friction in the formulas.


True true, and I have no idea how one would model that apart from calculating the area of the touching parts and fitting it with a single empirically determined parameter.


just when is the increase in friction higher, by increasing the bore (more circumfencial (sp?) area of the piston and also the cylinder) or the stroke (same area for the piston, but it rubs against more height (=area) of the cylinder)?


Good question. For the stroke increase case you get an area of 2pi * r * (S - s), while in the case of increasing the bore you have an increase of 2pi * (R - r) * s. Both are linear, but for increasing the bore the friction is higher for a shorter path, while in the stroke increase case the friction is unchanged but for a longer path (as you already pointed out).


as a side note: M = F * r only works if u dont have vectors. with vectors u gotta use M = r x F, since a x b = -(b x a)


Of course, what you say is true. And that is why I wrote: "the force on the piston times the orthogonal offset"... by orthogonal offset I meant the projection of the vector, reducing it to the even simpler non-cross-product formula. I could have been more clear on that.
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Deus ex Machina

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Post Fri Apr 22, 2011 8:01 am

Re: Bore & Stroke Discussion

to be honest id be quite happy if i understood that whole theoretical stuff (be it mechanics, physics or maths) in german, let alone english^^

i know enough to survive out there in this bad world, but the finer, or higher aspects elude me constantly, which can be frustrating, since i have some of this stuff @ uni and dont care for learning stuff till i can quote the whole book while not understanding even the first paragraph.
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svvander

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Post Fri Apr 22, 2011 7:12 pm

Re: Bore & Stroke Discussion

Idem ditto. I do understand all the theoretical basic stuff in Dutch, but the more advanced technical information in English seems like Chinese for me. Maybe if I read it more than 3 times.

For those who don't understand the basics of an engine, it could be easier to explain the basic principle of a 2-stroke engine.
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robt100

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Post Sat Apr 23, 2011 9:04 am

Re: Bore & Stroke Discussion

Another thing to take into account (although I know you said this was purely bore & Stroke) is the type and weight of materials used. Also, when does torque overtake power as the highest value (IE: Even an oversquare engine with more bore than stroke gets more torque than power most times). Also the stroke will affect rpm limits, I believe the basic explanation of power is torque x rpm (somehow?!) so then this also meets a trade off. And heavier materials being forced round at the same speed would have more momentum/turning force = torque. Dynamic compression (rather than static) would also play a big factor.
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Deus ex Machina

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Post Sat Apr 23, 2011 9:47 am

Re: Bore & Stroke Discussion

linear P (ower) = F * l (ength) / t (ime); meaning u pull something up 1 meter in a minute and the guy after you manages 10 meters in a minute => he had 10 times your power

length per time is speed (velocity).

=> P = F * v with P in Watt, F in Newton, v in m/s


now the crank shouldnt really travel in a straight line, but rotate.
So the imaginary point we moved in a straight line before with a speed of v now sits on the circumfence of the crank (outer edge of the "plate" thats part of the clutch) and also travels in circles.


circumfencial speed is PI*diameter (in meters)*RPM/60 (to get to the seconds)

v = PI*2*r*RPM/60, which we insert in the formula from the straight line

=> P (rot.) = r*F*PI*RPM*2/60

r*F = Torque; and we can get rid of the 2 and reduce the 60

=> P (rot.) = M*PI*RPM/30 in Watt;

=>P (rot.) = M*PI*RPM/30 000 with P in kW; M in Nm


as for the heavier mats:

the con rod connects to the piston in the center of the latter, which means every time the piston changes verlocity it gets pushed (upwards), pushes (downwards after the ignition) or gets pulled (intake phase) and all those forces attack in the center (or in case of after the ignition meet resistance there). That puts a significant stress on the piston as a whole since the outer edges dont want to go with the rest, so a weak piston would cause the con rod to make a dent in it middle, which in turn should cause the lower outer edge to bend inwards (and maybe the upper one too, so that the piston wouldnt fit the cylinder anymore...).

the heavier the material, the greater its resistance to changes in direction => has to be stronger too, which not all are, or would you make a piston out of gold? (Iron has the number 26 in the periodic table, gold 79, but pure gold is much more moldable)

another side effect of this increased resistance is the fact that it puts more force into the con rod (at the top point the piston doesnt want to slow down, so it stretches the con rod, while at the bottom it wants to continue downwards and therefore compresses the con rod); so you would have to make that one stronger too.

and all those forces continue onwards to the lower fixing of the con rod, the crank, so u gotta harden that one too.

now we have all those hardened and heavier parts, which all resist the increasing or decreasing power of the ignition if u change the amount off throttle. => laggy reaction to more (and less) throttle


and with higher RPMs the stress gets higher too, so you maybe need to introduce a artificial rev limit to keep the parts from failing



as for the higher momentum of the heavier parts: that only counts in a straight line, like running a car or a lorry into a wall, the car might be stopped, the lorry not. (and here we have some ppl who built their home directly in the firing line of the first curve of the country road after 5 km. at least once a year they have a lorry parking in the living room). Heavier stuff doesnt like acceleration (resists the ignition) or deacceleration which u only have at a downwards movement (piston) and there only in the second half, which translates to a quarter circle of the con rod @ the crank, and it pushing downwards when the con rod is at its lowest point isnt that healthy for the engine.


edit: now i know why theres a preview button -.-
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dread_darven

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Post Thu Apr 28, 2011 6:56 am

Re: Bore & Stroke Discussion

All those formulas give me a headache :P .
But testing things out, looking at the results and looking at paterns is my kind of thing, hehe

This stuff about bore and stroke got me going, i used an Engine analyzer (engine simulator) to do some science 8-) .
I did a couple of runs with bore and stroke 90/90 100/90 90/100 etc to look at the differences.
Here is the Excell file i put all the data in:
http://www.xs4all.nl/~brendel2/darc/Engine%20testing/Engine%20testing.xls
First two pages (source and backup) is for all the data and at the source page you can use the filter at the top row to just show brake hp or something else.
Also has some pages with just the data for bore stroke in HP and Tq etc.

Still abit raw and not user friendly but there is a load of information already in.
The program is pretty realistic if the data you put in is correct (the engine in my honda is almost completely on target)
The engine used is not an existing engine but something made up for easy manipulation as regards to bore and stroke increases.

Will put another excell file up when i done some more "science"
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Deus ex Machina

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Post Thu Apr 28, 2011 8:28 am

Re: Bore & Stroke Discussion

holy sh** numbers everywhere :-P

otherwise nice work
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dread_darven

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Post Thu Apr 28, 2011 8:55 am

Re: Bore & Stroke Discussion

It's indeed full of numbers :P .
But if your abit familiair in excel there is quite abit of information already in there which could be used to see alot of changes (not just HP and Tq numbers) in regards to bigger bore and longer stroke.
The program gives a buttload of information.... which i don't all understand :P .

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