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Let's talk about r/s

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17K views 25 replies 16 participants last post by  Import Builders  
A
#1 ·
First of all, lets define r/s ratio:

rod length / engine stroke

This makes people think that at a 1.5:1 ratio , the rod is only 50% longer then the crankshaft.

The problem with that is the crank throw is only half the engine stroke.

therefore it should be stated as:

rod length / crank throw

now this same 1.5:1 r/s ratio is 3:1. Sounds a little better doesn't it?

now lets get into angles. The "popular internet argument" is that a lower ratio radically increases rod angles, therefore pushing the piston into the cylinder wall.

Lets compare two popular engines.

first one being the k20, second k24, since it seems most people here are interested in the two.

the k20 has a 3.39 stroke, 1.7" crank throw.
the rod length is 5.985"

the rod to crank throw ratio is 3.52:1

now lets use some high school trig to figure it out. remember SOH CAH TOA?
lets use it to figure out the most extreme rod angle, at 3 oclock and 9 oclock.

the formula to use is cos(theta) = adjacent/hypoteneuse(spelling anyone?)

cosine is simply a button on your calculator(using a computer, hit "View" then
click "scientific".

theta is what we are trying to figure out(it will be the rod angle in degrees)

adjacent is the crank throw(NOT THE STROKE)

hyp is the rod length, center to center obviously

lets plug the #'s

cos(theta) = 1.695 / 5.473

cos(theta) = .3097 blah blah blah...

now "check" the inverse on the calculator and click cos, we have figured out a rod angle of 71.958 degrees.

now lets figure the k24

the stroke is 3.9" crank throw 1.95"

rod length is 5.985"

cos(theta) = 1.95 / 5.985"

cos(theta) = .3258

theta = 70.985

a whole degree.

just for fun lets toss in the f20, with its "awesome rod angle"

I couldn't find exact measurements, but i did find the r/s online at 1.82. We can just double that and inverse it. 1 / 3.64 = 2.74

theta is 74 degrees.

There you have it, 3 degrees between the f20 and the "terrible" k24.
 
#2 ·
many people think its only 50% longer because it IS 50% longer than the stroke. you wont hear people say "the rod is 3 times longer than half the stroke", which pretty much still means "the rod is 1.5 times longer than the stroke". key word: stroke, not crankshaft, which is why its called rod/stroke ratio.

in your case, it make sense to use half the stroke because you are trying to find the steepest angle which is when the crank creates imaginary right-angle triangle, but that's the only case when we will use the radius of the crank rotation. its simple just to make it rod stroke; why? 1:1 would mean the rod is the exact same length as the stroke; which would probally cause a collision but its possible to get away with that. now a 1:1 "rod/crank throw" would mean a definant collision. as you can see in your own calculations, the R/S ratio is simply a ratio; not a calculation for angle. changing it to Rod/Throw is just going to confuse lots of beginners.

also, the rod angle doesnt included the piston speed which is one of the reasons the k24 is under alot of stress when revved high. theres a calculator on SlowGT that shows the various loads and acceleration the engine sees throughout the crankshaft rotation at a given RPM.
 
A
#3 ·
any time you have a longer stroke engine, it will make power at lower rpms, period.

HP is what is important, not hitting a certain # of rpms.

besides, compare a k20 to a k24 with the same top end and everything else, you will find they make within a few % of hp at the same piston speed. Who cares if the k24 has a longer stroke. After this "merging" the k20 will be revving too many rpms, the k24 will completely take over.

actually R/S *is* a calculation for angle. take the r/s, double it(for crank throw, not stroke), and inverse it.

the adj/hyp part of the equation is essentialy a ratio, just calculated differently then usuall(its inverted)
 
R
#7 ·
The funny thing is that a LOWER r/s can provide some interesting benefits. It all depends on what you want to use the motor for.
 
1
#8 ·
I'm def no engine builder and starting to learn more about r/s ratios. I'm currently in the planning phase of my allmotor race engine build. Reliability is not key but still a factor. I've had an idea of using a K20 block with a 90mm x 102mm setup. From a r/s stand point this doesnt sound to great as far as what I'm learning but then again I've been told by a couple well know people that it sounds like it should make a good motor. Also, I'm trying to get the most out the motor and stay with in the guidelines of the NHRA's 2.6 liter. Was figured out for me that this setup would be 2595cc. To the best of my knowledge so far sounds like the motor will make good HP and torque down low and we'll have to dyno it to see about peak rpm's. Would like to hear comments good or bad and why you think that way.
 
G
#13 ·
I once worked out the effect of R/S on the dwell time. For a R/S change from 1.5 to 1.75 theres such as small effect on the dwell time its hard to see where it could change performance. Lets say at 12deg after TDC a R/S of 1.5 gives 2mm displacement from TDC. For a R/S of 1.75 you will get the 2mm after 13deg.

From what I understood of the advantages of a longer R/S, you can advance the ignition to achieve the same chamber pressures because of the longer dwell time. But by my calcs, you could only advance ignition by 1 deg. So for tuning an engine on pump gas even, where is the advantage in power by using a higher R/S?? (Obviously ignoring wear here)

My own opinion (after reading a lot of shit on the net) is that R/S just changes the piston speeds and accelerations, which changes the inlet and exhaust gas flows. This causes power to by won or lost.
 
D
#16 ·
well i would have to say its all about what u want to do becuz saying displacement is everything isnt so true what about a formula car i know they are way different motors but they have what v10's that are like 3 litres or sumthing like that. it all depends what racing u want to do. there is no perfect setup some thing can always break and nothing everything is built the same its more so on engine builder to help keep ur motor safe.
 
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#17 ·
duckmanEG said:
well i would have to say its all about what u want to do becuz saying displacement is everything isnt so true what about a formula car i know they are way different motors but they have what v10's that are like 3 litres or sumthing like that.
Click to expand...
actually F1 still does make power from displacement; a direct result from increased displacement is increased torque. F1 cars make almost the same amount of torque as other high-performance engines of the same displacement; but the fact that they are able to make that torque at a high-rpm is why they have lots of horsepower(remember TQ*RPM/5252=HP, to increase HP you can increase the RPM, the TQ, or both. Since FI was banned from F1, the TQ goes down, but theyve increased RPM to try to compensate). if you were to make the engine bigger but retain the high-rpm aspiration, the horsepower would increase significantly.
 
#18 ·
How about we redo the math with a little more precision & utilize significant digits.

all of my data is pulled from the tables on Arias's website

k20 stroke: 86.0mm, k20 rod length: 139mm (5.472") (rod/stroke = 1.62)
k24 stroke: 99.0mm, k24 rod length: 152mm (5.984") (rod/stroke = 1.54)

Since peak rod angles occur when the crank arm is perpendicular to piston travel, peak rod angles calculated using the following formula:

rod angle = cos^-1((stroke / 2) / rod length)

k20 rod angle = cos^-1((86.0mm / 2) / 139mm) = 71.97 deg = 72.0 deg
k24 rod angle = cos^-1((99.0mm / 2) / 152mm) = 70.99 deg = 71.0 deg

When significant digits are used, you come up with the same 1 degree difference, and it seems to favor the k24!

So how is it that a worse rod/stroke results in a seemingly more favorable rod to crank angle? The answer is you're looking at it from the wrong angle (excuse the pun). The rod angle is expressed as the angle above horizontal. So the closer to 90 degrees the rod angle is, the less piston sideloading you have. The k20's rod angle is closer to 90, so it will have less piston sideloading.

When you consider the s2000's engine parameters:

f20c stroke: 84.0mm, rod length: 153mm, (rod/stroke = 1.82)

f20c rod angle = cos^-1 ((84.0mm / 2) / 153mm) = 74.06 deg = 74.1 deg

So the difference seems to be only 4.1 degrees between the k24 and k20. Now to put that in perspective, let's consider the angle between the piston and the rod. (subtract crank to rod angle from 90)

k20 piston to rod angle: 18 deg
k24 piston to rod angle: 19 deg
f20c piston to rod angle: 16 deg

The k24 has a 19% greater angle between the wrist pin & rod than the f20c.
The k24 has a 6% greater angle between the wrist pin & rod than the k20.
The k20 has a 13% greater angle between the wrist pin & rod than the f20c.

The piston to rod angle gives you the most accurate glimpse at what the piston is trying to do. The piston wants to be at a 0 degree angle with the rod. The closer to 0 degrees you are, the better. Side loading does not increase linearly with respect to the piston to rod angle. I do not know the exact relationship, but the jump from 18 to 19 deg is worse than the jump from 17 to 18 deg.

So now I hope you can see that what was a seemingly insignificant 1 degree change in rod to crank angle is actually a 6% increase in piston to rod angle. Multiply that 6% PER REVOLUTION increase in stress and you start to see how quickly things can add up. Only 3 degrees separate the rod to crank angle of the s2000 from that of the k24, but those 3 degrees net a 19% change in piston to rod angle.

;)

I will add that displacement can make up for a LOT in terms of power. A higher displacement, lower rod/stroke motor can easily out power a lower displacement higher rod/stroke motor. It's only one part of the puzzle, but it gives you some very important clues as to how the engine will behave under stress.
 
#20 ·
allan5oh said:
You are correct, however rod angle at the wrist pin increases is not linear with piston side loading.

A difference between setup a at 5 degrees and setup b at 10 degrees isn't "double" the side loading.
Click to expand...
I mentioned that.

chunky` said:
The piston to rod angle gives you the most accurate glimpse at what the piston is trying to do. The piston wants to be at a 0 degree angle with the rod. The closer to 0 degrees you are, the better. Side loading does not increase linearly with respect to the piston to rod angle. I do not know the exact relationship, but the jump from 18 to 19 deg is worse than the jump from 17 to 18 deg.
Click to expand...
 
D
#22 ·
i believe that there is a point where r/s ratio becomes superior to displacement

k20 piston to rod angle: 18 deg
k24 piston to rod angle: 19 deg
f20c piston to rod angle: 16 deg

ok lets say that all these motors make 200 ft/lbs. of torque......

so 200 ft. lbs. is the resultant force (hypotenuse of triangle),
the opposite side of Theta is the Side load force,
the adjacent side of theta is the vertical force of the piston.

K24 sideload force = 65.1 ft.lbs / Vertical force = 189.1
k20 sideload force = 61.8 / Fv = 190.2
f20 sideload force = 55.1 / Fv = 192.3
i put the vertical force numbers there so u can check them with the pythagorean theorom A^2 + B^2 = C^2 (C being resultant force of 200 ft lbs.)

and remember, this is for 1 REVOLUTION!!!!!!!!!!!!

start multiplying those Sideload forces by revolutions PER SECOND and u will be surprised how much force ONE SIDE of a Cylinder wall will recieve in a second........ R/S ratio hype???? my ass
 
K
#23 ·
delslo9 said:
i believe that there is a point where r/s ratio becomes superior to displacement

k20 piston to rod angle: 18 deg
k24 piston to rod angle: 19 deg
f20c piston to rod angle: 16 deg

ok lets say that all these motors make 200 ft/lbs. of torque......

so 200 ft. lbs. is the resultant force (hypotenuse of triangle),
the opposite side of Theta is the Side load force,
the adjacent side of theta is the vertical force of the piston.

K24 sideload force = 65.1 ft.lbs / Vertical force = 189.1
k20 sideload force = 61.8 / Fv = 190.2
f20 sideload force = 55.1 / Fv = 192.3
i put the vertical force numbers there so u can check them with the pythagorean theorom A^2 + B^2 = C^2 (C being resultant force of 200 ft lbs.)

and remember, this is for 1 REVOLUTION!!!!!!!!!!!!

start multiplying those Sideload forces by revolutions PER SECOND and u will be surprised how much force ONE SIDE of a Cylinder wall will recieve in a second........ R/S ratio hype???? my ass
Click to expand...

honestly dude , fuck rs ratio i see every 1 revin 2.4s to 9k wtf , ohh well it is what it is . this has been discused so many times , do some searching.
 
I
#26 ·
A Ton of V8's have an erily similar deck height to a K20 block. Its incredible. Its like the exact same thing. Do you think they run large stroke and long rods? they have the same powerband as a honda. 8-10k rpm. I dont know one team running a long rod motor, and I know alot of them. And this is common through a gambit of classes.

These types of motors should be compared to Honda's becuase F1 and all that is not the same powerband and use that we are using our motors for, or even can build the same. So we have to compare to V8's who rev to about 10k. Good comparison.

They just have like 100+MM bores to most of them, so the displacement is through the roof. That would be the only drawback to us.

Long rod stroke ratio is purely a Honda import thing, but our import scene is over 20 years behind Domestics. Eventually people in imports will ignore trying to put the longest rod possible in a motor.

Jeff
 
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