B20 Vtec power. [Archive] - Ozhonda Forums

Terrano08

15-10-2009, 01:38 AM

You should be sitting around the 120kw mark.

I think you mean no ECU as in, no tune? I am very worried you're running on a B18 tune, the b20 obviously needs more fuel. 7200 is pretty premature also, but keeping in mind you havn't got any decent rod bolts holding the rods together. I'd get a decent build before you regret it.

yeah i meant its not tuned. in what way would i regret it ?? it actually goes pretty good for a minimal budget build. just cant afford internals so i gotta work on little things like, CAI, tune, and how to drive it properly..

should i shift every gear at redline or somewhere else??

Chr1s

15-10-2009, 12:48 PM

The small rod bolts of the standard b20 were not designed to see extended RPM the combination makes. They will fail, just give it time and it won't be an easy fix if it does.

You're short shifting the motor, I guarantee you that your power curve will continue to peak past 7200rpm yet you can't take advantage of it as the bottom end is not strong enough.

I'm not saying you can't spin it harder, or the moment you do a rod will decide to have a go at the bore, but it can happen.

string

15-10-2009, 02:06 PM

The small rod bolts of the standard b20 were not designed to see extended RPM and power the combination makes.
Why would increased power contribute to rod bolt failure? Extra downwards force on the piston doesn't make the bolts work harder to hold the cap on - the force is in the wrong direction.

GSi_PSi

15-10-2009, 02:42 PM

7200 is pretty gay lol.. if it is indeed stock bottom end (ie stock rod bolts), be very very carefull, otherwise thrash it and rebuild cause if your planning on revving out this motor its not going to last long.

Chr1s

16-10-2009, 10:52 AM

e240

16-10-2009, 11:05 AM

Why would increased power contribute to rod bolt failure? Extra downwards force on the piston doesn't make the bolts work harder to hold the cap on - the force is in the wrong direction.

I don't think that thats the correct way of looking at this though because it isn't just a push pull force but also a rotational force as the rod moves with the crankshaft. The top of the rod supports the piston, "added weight", the rod (and ultimately the bolts) are supporting this load whether the piston is going up or down - and there's never a "straight down" force applied except when the piston is TDC right? The rods are almost always at an angle.

tinkerbell

16-10-2009, 12:24 PM

string

16-10-2009, 12:39 PM

I don't think that thats the correct way of looking at this though because it isn't just a push pull force but also a rotational force as the rod moves with the crankshaft.

All of which can be represented by simple orthogonal forces.

- and there's never a "straight down" force applied except when the piston is TDC right? The rods are almost always at an angle.
The force from combustion is most certainly straight down onto the piston. From here it splits into two components; one directly down the rod at whichever angle it happens to be at. The other can be computed by vector subtraction, and is the force of the piston against the sleeve.

The top of the rod supports the piston, "added weight", the rod (and ultimately the bolts) are supporting this load whether the piston is going up or down

There are two elements of rod load:
1) Inertia property of the piston/rod. Changing the direction of the mass requires acceleration, which requires force. When the piston is accelerating downwards (upper half of the stroke) the rod is in tension. When the piston is accelerating upwards (lower half of the stroke) the rod is in compression.
2) Combustion forces. Always compressive on the rod and has a the majority of it's power in the upper half of the stroke.

The rod bolt must be able to hold the rod cap on under the extreme tension stress of the rod. When the rod is under net compression, the rod bolts don't have anything to fight - the rod-cap bearing is completely unloaded.

Conclusion: Combustion forces are in the wrong direction to have any effect on the rod bolts.

string

16-10-2009, 12:44 PM

the force never changes direction does it?

As long as we're both talking about internal combustion 4 stroke motors, I am quite certain that combustion forces are always downwards.

or the caps and bolts are weightless and don't have inertia?

Tensile bolt stretch from cap and bolt mass changing direction is a function of piston acceleration - which is independant of engine power output. Since superposition is in full swing, I can say that when talking about combustion forces, it's irrelevent.

now i have heard everything!
A stunning jibe, but no substitution for intelligence.

tinkerbell

16-10-2009, 12:51 PM

re-read what you quoted, then think of way to explain why "increased power" must only relate to downward forces created by combustion.

you are just being silly.

string

16-10-2009, 12:56 PM

If you have an argument, let it out - don't beat around the bush making me guess what you're thinking.

The context is a stock rev-limit B20 block. Increase power without adding more revs? Add torque; a synonym for combustion force.

tinkerbell

16-10-2009, 01:00 PM

nah, you win,

or something like that...

string

16-10-2009, 01:07 PM

So is there any reason you came into this thread other than to block it up with claims you won't substantiate, or to add some childish manner?

Chr1s

16-10-2009, 01:16 PM

I think string means the main reason for rod bolt failure is the tensile forces implied are alot more to consider than a bolt which performs better in shear stress.

edit: ffs, just missed that whole load of crap above.

tinkerbell

16-10-2009, 01:22 PM

edit: ffs, just missed that whole load of crap above.

yeah, the moral of the story was that "increased power" doesn't have any physical effects on any part the bottom-end beside the top of the piston, the piston pin, the rod (but not the rod cap) the top half of a rod bearing and the top half of the rod journal, the bottom half of the crank bearing and the main cap...

string

16-10-2009, 01:26 PM

My point is that from a geometric perspective, combustion forces cannot cause tensile rod forces.

yeah, the moral of the story was that "increased power" doesn't have any physical effects on any part the bottom-end beside the top of the piston, the piston pin, the rod (but not the rod cap) the top half of a rod bearing and the top half of the rod journal, the bottom half of the crank bearing and the main cap...

Are you going to have a discussion like an adult now? A condescending attitude does the masses few favours in an academic debate.

tinkerbell

16-10-2009, 01:29 PM

My point is that from a geometric perspective, combustion forces cannot cause tensile rod forces.

i agree.

but your comments don't really assist do they?

they stop at the first part of a four stroke cycle...

string

16-10-2009, 01:31 PM

i agree.

but your comments don't really assist do they?

they stop at the first part of a four stroke cycle...

We're comparing an Engine with a slightly more powerful version of itself - the differences between the two stop after the first part of a four stroke cycle...

To make myself crystal clear: The context of my argument stems from the combustion force as the variable in question. If you want to discuss how increased power affects more than the power stroke, be my guest (though if this were the case, I'm sure you would have raised it in post #12).

Chr1s

16-10-2009, 01:50 PM

yeah, the moral of the story was that "increased power" doesn't have any physical effects on any part the bottom-end beside the top of the piston, the piston pin, the rod (but not the rod cap) the top half of a rod bearing and the top half of the rod journal, the bottom half of the crank bearing and the main cap...

I beg to differ, manufacturers, or should I say engineers, are more inclined to investigate the tensile forces of the rod bolts. If you mean incease in HP with a result in twisting as such that could cause shear across the bolt between the cap, well thats another story - and how often does this happen before the rod gives up?

GSi_PSi

16-10-2009, 01:52 PM

why dont use stop the arguement lol b20 stock rod bolts are shit and will fail when you want to rev more

Chr1s

16-10-2009, 01:54 PM

It's a good discussion Sam, well heated between tinker and sting :zip:

tinkerbell

16-10-2009, 01:57 PM

why dont use stop the arguement lol b20 stock rod bolts are shit and will fail when you want to rev more

what about if you don't rev MORE,

but rev HARDER? i.e. due to more torque? (i.e. being created by higher combustion pressure)

will the rods (and bolts) suffer higher forces then?

string

16-10-2009, 01:58 PM

why dont use stop the arguement lol b20 stock rod bolts are shit and will fail when you want to rev more
True story. The primary factor is how high you want to rev. The force required to reverse the direction of of the piston and rod mass at TDC grows exponentially as you rev higher - and it all goes through the rod bolts.

Chr1s

16-10-2009, 02:25 PM

what about if you don't rev MORE,

but rev HARDER? i.e. due to more torque? (i.e. being created by higher combustion pressure)

will the rods (and bolts) suffer higher forces then?

Tinkerbell, is there another way to explain the theories, I still can't see how power has a direct relationship to the stresses on the rod bolts as opposed to RPM.

tinkerbell

16-10-2009, 02:31 PM

fatboyz39

16-10-2009, 02:38 PM

hey ive got a dc2 vtir with a stock engine conversion to B20B8 block and vtir head. The engine revs out to 7200 and no work has been done at all. no cold air intake, no tune, NOTHING. anyone know what kind of power this should be making and ide also like to know what revs to shift at or would anyone by any chance have a dyno sheet for this ??

Strongly advise you change your rod bolts.

string

16-10-2009, 02:52 PM

imagine a nitrous B20B.

stock RPM limit.

combustion goes bang, rod goes down.

where are the other three rods?

there is another piston/rod going down too, but it is being pulled down by the crank with no force on top.... (NB - this rod and piston has inertia)

imagine what is happening to the rod bolts on that rod?

are the forces on those rod bolts going to be the same if the forces on the other piston are higher due to the higher combustion pressure?

If the forces pulling the other rod down were somehow greater with the higher power, then you would be accelerating the piston faster (since mass is obviously conserved) - which of course does not happen in a state of constant engine speed.

The forces on the remaining non-powered rods are determined by the angular speed of the engine, not the torque onto the crank-shaft. For a given engine speed: the physical nature of the rod and piston assembly determines how much force is required to overcome it's state; the reverse is not true: increased torque on the crank pushes the piston harder. If you want to balance the equation, increase the "load" on the engine - which involves increasing the angular acceleration of the load - the exact expected result!

The forces on the rod can be examined one at a time and then summed to get the final result. Comparing two otherwise identical engines at the same operating speed means you can cancel out the inertial loads as they are equal in each case.

Summary: You can push an object a given distance in a given time with only one constant force. You can't push twice as hard and have it take just as long. (I.e. you can't push/pull a rod harder yet have it move at the same speed).

tinkerbell

16-10-2009, 03:29 PM

If the forces pulling the other rod down were somehow greater with the higher power, then you would be accelerating the piston faster (since mass is obviously conserved)

of course. that is why the car with more power (force) accelerates faster!

- which of course does not happen in a state of constant engine speed.

way to change the subject.

we (i assume everyone except you) are talking about an engine in dynamic use, for example drag racing, not sitting on a constant 2389rpm...

The forces on the remaining non-powered rods are determined by the angular speed of the engine, not the torque onto the crank-shaft.

yep, at a constant speed, lol!

For a given engine speed: the physical nature of the rod and piston assembly determines how much force is required to overcome it's state; the reverse is not true: increased torque on the crank pushes the piston harder. If you want to balance the equation, increase the "load" on the engine - which involves increasing the angular acceleration of the load - the exact expected result!

yep, again at a constant engine speed...

The forces on the rod can be examined one at a time and then summed to get the final result. Comparing two otherwise identical engines at the same operating speed means you can cancel out the inertial loads as they are equal in each case.

yep, exactly, at a constant engine speed.

Summary: You can push an object a given distance in a given time with only one constant force. You can't push twice as hard and have it take just as long. (I.e. you can't push/pull a rod harder yet have it move at the same speed).

yep, that is why the nitrous one accelerates faster :thumbsup:

(and why the forces on the rod bolts are higher...)

you cant have an engine accelerate to 7200rpm faster than another (identical) once without the forces on all the parts being higher, can you?

string

16-10-2009, 04:04 PM

The forces resulting from "inertia" are completely and only dependant on the rotating speed of the engine. A 500hp B20 at Xrpm has the same "inertial rod loads" as a 200hp B20 a the same Xrpm.

The power stroke of one piston only affects the single rod, and the engine load. If you simultaneously increase the torque by 10% but also increase the load by 10% - acceleration remains identical, and the "inertial rod loads" curve vs time remains identical.

It doesn't matter how fast the engine is accelerating, the loads on the rod from the mass of the piston is dependant only on the speed at which the crank is rotating.

When the crank attempts to move the pistons/rods upwards in their bores, it asks them "how much force do I need to give you such that you move at the same rate that I do" - the rods answer and only the required force is applied - no more, no less.

If you agree that the only tension [rod bolt related] stress results from the piston/rod's mass, then you are agreeing that the rod bolt's stress is dependant only on the rotation speed of the engine.

you cant have an engine accelerate to 7200rpm faster than another (identical) once without the forces on all the parts being higher, can you?
If the engine is accelerating faster because it has less load, then it will accelerate without more force on any part.
If the engine is accelerating faster due to more torque then the difference will be the compression rod stress resulting from increased cylinder pressure. Because of the shorter time scale, you'll reach the peak 7200rpm "inertial rod loads" faster but the peak rod tension loads will be identical.

Summary: Revving harder doesn't pull the 'idle' rods down harder - revving higher does that.

tinkerbell

16-10-2009, 05:54 PM

If the engine is accelerating faster due to more torque then the difference will be the compression rod stress resulting from increased cylinder pressure. Because of the shorter time scale, you'll reach the peak 7200rpm "inertial rod loads" faster but the peak rod tension loads will be identical.

i am sorry, it is hard to follow what you mean here?

string

16-10-2009, 06:12 PM

i am sorry, it is hard to follow what you mean here?

I mean that a faster acceleration doesn't mean greater force on the rods from the mass of the piston - it means the forces grow to the same size, faster.

When I say the "inertial rod loads" are dependent on the engine speed, I mean that the engine speed can be anything. It can be changing, or constant.

A higher torque otherwise identical engine transmits a greater torque to the load thus it's angular acceleration will be greater. The "inertial rod loads" are exactly the same as a function of engine speed, just that the engine speed builds up quicker.

All the forces can't be the same otherwise the result would be the same. The forces arising from the mass of the rotating assemblies will be identical in magnitude however, regardless of the engine's angular acceleration. Since the time from 0-MAX rpm will be less, the power absorbed/transmitted by/through the various means, will be larger since you have reduced the time to do the same work.

ewendc2r

16-10-2009, 07:29 PM

I hate to say it, but the comments about your condescending approach to 'offering opinion' is spot on, there isn't a need for people to act god, if a question is asked it'd be great if a reasonable informative response was posted..

I only say this because you've done it in response to me before, and I couldn't figure out in what way it could be construed as positive input..

Oh, and I'm pretty sure (after some thought) that string is technically spot on.

/breathe.

joyride

17-10-2009, 06:50 AM

i didnt read all of the above because its way too technical for me.

tinkerbell/string: i know both of you guys have a wealth of knowledge and experience; its good that you are both trying to come to the same conclusion but please do not try and argue with each other directly. you know PM is always available to discuss it there.

cheers guys. its good to see OH still has some 'thinkers' onboard :)

Benson

17-10-2009, 08:27 AM

So why do people run forged rods in their turbo motors with >15psi through the motor and cap the revs to 7200rpm? i.e EVO's

IF what your saying is right string, then people dont need forged rods

aaronng

17-10-2009, 08:54 AM

You guys forgot that compression and elongation forces on the rod bolt affect the metal differently. As long as your are generating more combustion force than stock, that means the other piston in the intake cycle would experience a stronger elongation force when it is being pulled down by the crankshaft. The metal in the rod bolt can eventually experience fatigue failure after sufficient cycles if the force exceeds the amount needed to slightly stretch the metal in the rod bolt.

tinkerbell

17-10-2009, 11:59 AM

string

17-10-2009, 06:56 PM

So why do people run forged rods in their turbo motors with >15psi through the motor and cap the revs to 7200rpm? i.e EVO's

IF what your saying is right string, then people dont need forged rods

I haven't discussed the strength of the rods in compression because I'm not qualified to make comments - I don't know anything about the specific materials in question. You can't avoid a higher torque combustion event from pushing down on the crankshaft (through the connecting rod) harder - if you're making more power than a stock cast rod can handle in compression (or handle in the event of detonation) then you'll have to go for a stronger item.

As long as your are generating more combustion force than stock, that means the other piston in the intake cycle would experience a stronger elongation force when it is being pulled down by the crankshaft.

Consider two engines, one 10% more torquey than the other but otherwise identical. They are both connected to a load which holds the engine speed constant (one load is 10% larger obviously). One engine is putting out 10% more power but both are operating at the exact same speed.

In both cases, the rods and pistons are moving at identical rates. If two objects have equal positions at all points in time, they also have equal velocities and equal accelerations. Since they have equal accelerations, we know that from F=ma, they must both be receiving the exact same force, even though one engine is producing 10% more power than the other.

aaronng

17-10-2009, 08:01 PM

Consider two engines, one 10% more torquey than the other but otherwise identical. They are both connected to a load which holds the engine speed constant (one load is 10% larger obviously). One engine is putting out 10% more power but both are operating at the exact same speed.

In both cases, the rods and pistons are moving at identical rates. If two objects have equal positions at all points in time, they also have equal velocities and equal accelerations. Since they have equal accelerations, we know that from F=ma, they must both be receiving the exact same force, even though one engine is producing 10% more power than the other.
If both engines are at the same speed, and both have equal acceleration and hence equal force, then both engines are producing the same amount of power. The more "powerful" engine is however using less throttle to keep the engine spinning at the same speed as the weaker engine.

string

17-10-2009, 08:37 PM

If both engines are at the same speed, and both have equal acceleration and hence equal force, then both engines are producing the same amount of power. The more "powerful" engine is however using less throttle to keep the engine spinning at the same speed as the weaker engine.
Both are at the same speed and thus have equal piston accelerations. One engine has 10% more torque than the other and neither has angular acceleration - of course they're not producing the same power.

Both engines can effectively be thought to be at full throttle - the speeds of both engines are mediated by the choice of load (10% larger to counter 10% more power).

aaronng

17-10-2009, 10:14 PM

Both are at the same speed and thus have equal piston accelerations. One engine has 10% more torque than the other and neither has angular acceleration - of course they're not producing the same power.

Both engines can effectively be thought to be at full throttle - the speeds of both engines are mediated by the choice of load (10% larger to counter 10% more power).

Torque = Force x Vector distance

where Vector distance is the horizontal distance between the point at which you are measuring engine torque (crankshaft) and the point where the force is applied (piston-rod pin)

Since we assume that the engines are identical, then the horizontal axis distance from pin to crankshaft is the same for both engines and the force is the same for both engines, hence the torque must also be the same.

string

17-10-2009, 10:23 PM

Since we assume that the engines are identical, then the horizontal axis distance from piston to crankshaft is the same for both engines and the force is the same for both engines, hence the torque must also be the same.
The force on the combustion stroke piston is NOT the same. The example states that one engine has 10% more torque than the other = 10% more force down on the combustion stroke piston.

The remaining 3 pistons (who's motion is determined entirely by the crankshaft, unlike the combustion piston which has an unbalanced [without a load] force) and rods of both engines share the same position function, and thus velocities, accelerations and therefore forces from the crankshaft.

aaronng

18-10-2009, 10:56 AM

The force on the combustion stroke piston is NOT the same. The example states that one engine has 10% more torque than the other = 10% more force down on the combustion stroke piston.

The remaining 3 pistons (who's motion is determined entirely by the crankshaft, unlike the combustion piston which has an unbalanced [without a load] force) and rods of both engines share the same position function, and thus velocities, accelerations and therefore forces from the crankshaft.
So if you say that the 3 remaining pistons have motion determine by the crankshaft, and one engine's combustion piston is making 10% more force, then how can you assume that the other 3 pistons have the same force as the weaker engine? The crankshaft is directly driven by the combustion piston, so if you have 10% more force, that will also exert 10% more force on the other 3 pistons.

P = T x RPM / 5252

If you keep the two engines at the same constant velocity (constant RPM), then both engines would be making the same torque at the same RPM, otherwise one engine would have more power and accelerate more than the other engine (as per above equation). Even if you had both engines with identical velocities and accelerating at the same rate (increasing RPM), then both engines would also have to be exerting the same amount of torque, assuming that both engines are identical in terms of friction/losses because power has to be identical between both engines so that acceleration is identical.

F = m x a
T = F x r
Substituting gives T = m x a x r

Therefore if you assume that m (mass), a (acceleration) and r (horizontal vector) are identical for both engines, then T (torque) must be the same as well. If you say one engine has 10% more torque, then that violates the 2nd law of motion since the force has to be 10% higher as well.

string

18-10-2009, 11:31 AM

P = T x RPM / 5252

If you keep the two engines at the same constant velocity (constant RPM), then both engines would be making the same torque at the same RPM, otherwise one engine would have more power

Exactly... One engine does "have more power" than the other - the entire context of this discussion, glad you're with it.

and accelerate more than the other engine (as per above equation).

The 10% more powerful engine has a 10% larger load. The loads absorb energy such that both engines are at constant angular velocity. Read the example again.

Even if you had both engines with identical velocities and accelerating at the same rate (increasing RPM), then both engines would also have to be exerting the same amount of torque, assuming that both engines are identical in terms of friction/losses because power has to be identical between both engines so that acceleration is identical.

Again, the loads are not identical. The 10% more powerful engine is flowing 10% more power to the load. Angular acceleration is zero in both cases.

F = m x a
T = F x r
Substituting gives T = m x a x r

Therefore if you assume that m (mass), a (acceleration) and r (horizontal vector) are identical for both engines, then T (torque) must be the same as well. If you say one engine has 10% more torque, then that violates the 2nd law of motion since the force has to be 10% higher as well.
The torque at the load is 10% higher with the more powerful engine. Since the magnitude of the load is 10% larger (as cited in the example), it is in the same steady state as the smaller load. Disregarding your bastardization of the equations, the force is 10% larger - on the combustion rod and on the load.

So if you say that the 3 remaining pistons have motion determine by the crankshaft, and one engine's combustion piston is making 10% more force, then how can you assume that the other 3 pistons have the same force as the weaker engine?

You have two engines spinning at the same rate. One engine is producing 10% more power - it's combustion events give a 10% greater net torque on the crank. There is no angular acceleration of the engine - the remaining 3 pistons have motion determined entirely by the crank - the crankshafts in both cases must be providing the exact same force. The only way for the forces to be different in the more powerful engine is if it were spinning at a different rate - which it isn't.

The crankshaft is directly driven by the combustion piston, so if you have 10% more force, that will also exert 10% more force on the other 3 pistons.

Incorrect. The load will be driven with 10% more force - the pistons require only as much force as necessary to facilitate their motion - a motion which is identical to the weaker engine.

The only way to increase the force on the remaining 3 pistons is to increase the angular velocity of the engine. I specifically crafted this example with equal engine speed, but non-equal power flow to gracefully simplify the concept.

aaronng

18-10-2009, 11:44 AM

The 10% more powerful engine has a 10% larger load. The loads absorb energy such that both engines are at constant angular velocity. Read the example again.

Ahhh, I missed that 10% load increase that you posted earlier. My bad. Thanks for pointing that out to me again.

Back to the stress on the rod bolt, my understanding is that it is based on the acceleration of the piston/rod being pulled down by the crankshaft. F = m x a, so the additional acceleration is from the increased force which can be said to come from the engine's additional torque output. So if someone frequently enjoys full throttle short 1st gear acceleration, the rod bolts are subject to fatigue quicker than someone else with an ultra torquey engine but rarely uses very quick engine RPM increases. I guess you could say that a commo V8 doesn't snap rod bolts eventhough it has 3 times the torque of a b20vtec because the acceleration of the engine components is kept low by using higher gearbox ratios.

string

18-10-2009, 11:53 AM

Back to the stress on the rod bolt, my understanding is that it is based on the acceleration of the piston/rod being pulled down by the crankshaft. F = m x a, so the additional acceleration is from the increased force which can be said to come from the engine's additional torque output. So if someone frequently enjoys full throttle short 1st gear acceleration, the rod bolts are subject to fatigue quicker than someone else with an ultra torquey engine but rarely uses very quick engine RPM increases. I guess you could say that a commo V8 doesn't snap rod bolts eventhough it has 3 times the torque of a b20vtec because the acceleration of the engine components is kept low by using higher gearbox ratios.

If you were to blast quickly through 1st gear (full throttle, or you have a 10% more torquey engine for example), the rod bolt stresses and piston accelerations rise to the same levels, but more quickly. Their values remain identical with respect to the engine speed; engine speed however, now grows faster with respect to time.

The rotating assembly (including the load) is one connected unit. If your load offers 1000 times the resistance of pulling the piston down, it receives 1000 times the force. A V8 with 3 times the torque offers it to the load - if this triple torque happens to accelerate the load quickly, then the rod bolts motion is "accelerated" - but the rod bolts/rods/pistons are always accelerating up and down in their bores. Engine angular acceleration results in a growth in the piston acceleration - the derivative of acceleration, jerk. Newton gave us force = mass x acceleration with no mention of jerk.

The only way to keep the engine component (rod/pistons) acceleration low is to keep the engine speed low. Engine angular acceleration results in engine component jerk.

Engine angular acceleration is non-issue - unless you happen to know something I don't about high tensile steel's behavior in jerk.

Terrano08

19-10-2009, 11:55 PM

so listen, lets say i dont want to change my rod bolts coz im a lazy **** !! when should i be shifting gears ??? and what little things (not internals) can i do to enhance power??

string

20-10-2009, 12:00 AM

Don't exceed the stock rev-limit of the B20: 7000-7200. Stories of LSVTECs with spun rod bearings 3 months down the track are plentiful. Rev high on the stock bolts and you will regret it. The difference in forces between 7000rpm and 8000rpm is much bigger than the numbers alone might suggest.

The usual advice is as follows: Breathing mods (intake/header/exhaust), then tuning to make the most of what you have. A stock ECU is doing you no favours.

IIRC there is a stock block (high comp) B20VTEC in the Build forum with bolt-ons and tuning - ~125kW. Sounds good to me...

Lerlo

21-10-2009, 03:55 PM

I heard of issues of Con-rod stretch as well. Is the stock B20 conrods able to handle, say 8500-9000 rpm, without the piston hitting the head?

Terrano08

22-10-2009, 11:11 AM

Just wondering what type of bolts they are, are they a dingle bolt with aa thread on the other side of the rod or are theynut and bolt. And what is the tightneing torque required? thanks

lithium

22-10-2009, 12:00 PM

this is an interesting discussion guys, i learnt a lot from it :)
can i try to summarize, is this right?

increasing the peak torque output from an engine will result in a greater maximum compressive force on the rod of the piston on the power stroke. (this is why boosted evos et al. use forged rods to prevent rod failure due to excess compression). however increasing peak torque has no effect on the max force acting on the rod bolt.

on the other hand increasing the maximum engine speed increases the max force applied to the rod bolts. increasing max engine speed has no affect on the maximum compressive force on the rod.

so if you want to raise your rev limiter without increasing peak torque you only need to change the rod bolts. however if you increase peak torque you need to consider changing the rods themselves.

(in practice i don't think anyone changes rods without also uprating the rod bolts while they have the engine apart!)

GSi_PSi

22-10-2009, 01:00 PM

Just wondering what type of bolts they are, are they a dingle bolt with aa thread on the other side of the rod or are theynut and bolt. And what is the tightneing torque required? thanks

ARP ROD BOLTS - 208-6001,. Tightening instructions will come with it.
As with molylube etc

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22-10-2009, 01:10 PM

The torque requirement is more than the stock rod bolts (more clamping force). This means that when you install them on the stock rods, the big end (the actual rod material) is compressed ever so slightly more. The internet seems to think this is a problem which requires remachining of the big end bore - YMMV.

on the other hand increasing the maximum engine speed increases the max force applied to the rod bolts. increasing max engine speed has no affect on the maximum compressive force on the rod.

Increasing engine speed affects the compressive stress of the rods. The pistons are accelerated both downwards (rod tension) and upwards (rod compression).

Is the increase enough to warrant a forged rod? I don't know. Seems to be plenty of builds out there using stock LS rods with uprated bolts whilst revving sky high.