overZealous1

iTrader: (27)

i would like to see a stock block free rev to 10k, lol. think the stock rod bolts would let loose. most of the posts here are right in their own perspective though. rpm is mostly dependant on rod length/strength and the valetrain keeping a valve off of a piston. many other factors but for the most part this is the main concerns.
ya know what, i just got back from skiing and i don't have the energy to recap and say everything that has been said again in my own words, lol.

Enron Exec

You need to be more precise about your question. Exactly what inernals are we looking at when concerning stress? It could be a number of bearins, the pistons, the connecting rods, the head gasket, etc.

First off, like some ppl already mentioned, a FI VQ making peak hp at 4300 rpms has some serious problems, assuming we are sticking to your 400whp/400tq example. You will see why if you look carefully at a few FI VQ35DE dynos. Torque and horsepower cross on the dyno at 5252 rpms on all dynos. This is true of all motors. This might seem like some sort of anomaly, but if you carefully look at the relation between power and torque, you should see why.

Torque = (Hp X 5252) / RPMs

5252 is simply a constant for the conversion of power and torque. Power and Torque are directly related. Dynos calculate power by multipling the torque, measured at the hub or the wheel of a car, by the rpms the motor is spinning. Use a little middle school algebra and you get this equation from the one above.

Hp = (Torque X RPMs) / 5252

Back to the question. Since connecting rods seem to be a questionable internal in the VQ, i'll assume that is what we should talk about. In your example the car making peak TORQUE at the lowest rpm would be putting the most stress on the CONNECTING RODS with all else being equal. Your calculations are correct, but more accurately said, torque is what breaks things. Torque is a unit for rotational force. Force acting on an object is what causes things to move. Forces acting on an object in the proper directions and porportions will cause things to break. Basicly, a motor making 400 foot pounds of torque at 2000 rpms would have to put twice the force on each connecting rod, per stroke, as compared to a motor making the same 400 ft-lbs of torque at 4000 rpms because the 2000 rpm motor had half the combustion strokes as the 4000 rpm motor. We could get into more detail such as rod angle at peak cylinder pressure but...

mraturbo

iTrader: (1)

"Torque and horsepower cross on the dyno at 5252 rpms on all dynos. This is true of all motors. This might seem like some sort of anomaly, but if you carefully look at the relation between power and torque, you should see why."

Only if the they are on the same scales. Otherwise they can cross or make peak power before or after.

Michael

VRT
www.ViolentRacing.com

AmyCroft

iTrader: (3)

But not by definition - SAE standard

Cheers Amy -

Quamen

iTrader: (14)

350z dyno graph

Resolute

iTrader: (2)

I think this where people are getting confused, as I think Sharif had the same misconceptions. First, you accurately described the relationship between hp and tq, but not the effects. Certainly, hp is a relationship between FORCE and TIME that measures the amount of WORK being done. This leads to all manner of discussions on how to make horsepower and which is more important and how best to increase tq to make hp, or rpm to make hp, etc.... So for all those looking to bring up some of these arguments or ideas, it is not important for the question of which of our two engines puts thhe most stress on its reciprocating assembly. What is important in what EnronExec posted is to note that WORK (HP in our case) is a dependent calculation of the relationship between two actual measurements- FORCE (torque in our case) and TIME (rpm in this case).
Now having said that, it is easy to note that with our engines, at a lower rpm more work is being performed per each stroke than at a higher rpm to achieve the same result. This might seem like the lower rpm engine is making more "bang" at 4300rpm than the 6100rpm engine, and therefore more load on the piston, but that is incorrect. Remember, these are hp figures, and hp is the work being performed, and is not the load being applied. The "bang" is the actual FORCE being applied, and therefore the measure of load on the piston, and the force being applied is measured as TORQUE. I can guarantee you that in either rpm the max torque and max hp will not occur at the same value, and the equation EnronExec posted that shows the relationship between hp and tq, is good to prove exactly that if you want to play around with it.
So, we know that the hp peaks at diferent rpm, and that hp means squat in regards to loads on the piston, as it is only a calculation of work being performed and not the amount of force being applied. The loads are increased with torque, because it is when the max torque is applied that the "bang" on the piston is the hardest, and as previously discussed, the loads increase with rpm. In either engine, the hp figures help describe the relationship on how each is achieving the same amount of work, but the max torque (or max "bang" on the piston) is also the same for each engine, and therefore the load on each engine is identical and independent of how soon or late in the rpm the engine does most of its work. The only factor between these two engines that has any difference on piston loads, and therefore stress, is the difference in rpm. The 6100rpm engine must rev higher to achieve the same amount of work(hp) and therefore induces more load than the engine that only needs to rev to 4300rpm.
Also, with steel, compressive loads do not induce fatigue stress on the conrods, only tensil loads do, and is another reason the rpm will place more stress on the engine than the max torque being applied.
And sorry to those who wanted that link, I am still looking, but you might have to go to SAE's sitet to get it, and it costs. If i can figure out which cd I burned it to, I'll post it, I promise.
Will
EDIT: because i am better at math than spelling

overZealous1

iTrader: (27)

i believe from the original post he is eliminating most engine theories and looking at a hypothetical idea. based on a time function (rpm) the motor generating the same power at the lower rpm will be intaking more air/fuel per piston firing than another motor generating the same power at a higher rpm.
it is the main basis for not firing N2O at a low rpm. the N2O will inject the same amount of mixture whether at 1000rpm or 5000rpm, thus creating a huge amount of stress at lower rpms. think of the motor as a string chopping machine. the slower it chops (valve opening/closing), the longer the string (more a/f into cylinder). ya get the idea. again this is only true for N2O injection where it is independant of flow rates or cam profiles or rpm to make the power and is only based on jet size on a constant time scale.
i believe this was his idea from the original post. and yes it is true for N2O injection.

prescience

I agree with most of what resolute says but one thing I haven't worked out is why torque (bang as you call it) falls off as the revs rise beyond its peak value; I mean the piston is still being pushed down by the compression in the cylinder and the load would seem to be the same.

Perhaps its explained in the text above, but I can't find the answer right now

Resolute

iTrader: (2)

This part is grossly incorrect. The amount of air/fuel ingested is not greatest when the motor makes the most power, it is the greatest when the motor makes the most torque. In our example of two identical engines making the same power and torque, only at different rpms, the amount of air ingested at the point when either engine makes its identical 400 lb/ft of torque will be the same, and therefore the compression loads occured will be identical, albeit at different rpm.
The loads placed on the engine has NOTHING to do with when it makes its peak hp, it is a result of when and how much peak torque is applied and the top speed of the engine.

Resolute

iTrader: (2)

Imagine our hypothetical engine revs to a whopping 10rpm. It has one cylinder and to make things even easier is exactly one liter of discplacement. Our super liter engine has one intake valve and one exhaust valve, and makes one hp at that screaming 10rpm redline. At one rpm the engine ingests some air, ignites it, and then blows it out and moves on to the next cycle. One rpm is fun and all, but we want performance so we rev it up to two rpm. At two rpm more air/fuel is sucked in than at one rpm, it is compressed, and then exploded and pushed out. Now we noticed something interesting at two rpm- because more air was inhaled than at one rpm, we made slightly more torque at two rpm than at one rpm. This is because with more air/fuel to blow up, we had more energy to push on the piston. Well, we know this super liter engine discplaces one liter of air in theory, so we measure and find out that at two rpm it only sucked in .5 liter of air/fuel mixture. At one rpm it was even less. Being the performance freaks we are, we won't settle for that and throttle this mean machine to 5 rpm. At 5 rpm our little super liter engine is now inhaling one full liter of air and fuel to burn, and as a result, it is exploding and pushing the piston down much harder than at 2 rpm. This means we have more torque at 5 rpm than at 2. Now it gets interesting. We really go nuts and rev this puppy to 7 rpm. We notice at this rpm there is less air/fuel ingested than before, and therefore less torque. This means that the engine breathes its best, and therefore makes the most torque (or produces its biggest bang) at 5 rpm. Because it takes in one liter of air at 5 rpm and the cylinder physically discplaces one liter of space, at this rpm the engine is breathing perfectly. this is called the Volumetric Efficiency (Ev) of the engine, and at 5 rpm our super liter engine has an Ev of 100%. This is very good. No wonder its called the super liter engine. At any other rpm the Ev is less than 100%, and is why at any other rpm, the torque will be less. The interesting part is that the power of our engine is still climbing. At 7 rpm the torque is less, as the engine is drawing in less air/fuel to burn, but power is still up. The reason is because power is, again, a measure of WORK, and not a measure of the actual "power" pushing on the piston. The engine, in other words, is still producing work. This is because the amount of work performed is also a function of rpm AND torque. The force pushing the piston is less, but it is still pushing more often. Imagine loading a 100lbs into your car. You can do it all at once or you can load 50lbs twice. The amount of work done is the same, but one required more force and less time, the other half the force but twice the time. HP is WORK, TORQUE is FORCE, and TIME is RPM. Just as EnronExec posted, HP = Torque * RPM So even though torque is falling off, it is not falling off faster than the rpm is climbing. Same effect as lifting 50lbs twice in the same amount of time you lifted 100lbs once. Hang on to your hats, because we're going to redline the super liter at 10 rpm. At 10 rpm, our engine makes less torque than at 5 rpm, so we know it is not intaking as much air/fuel, and therefore does not have the greatest "push" on the piston at this rpm, but the power tops out at this rpm because even with less torque, the engine is still making enough revolutions to compensate. Any faster, and the lack of air/fuel begins to take its toll and besides less torque, less hp is made. It would be like lifting 20lbs only 4 times to load your car instead of the 50lbs twice. The amount of work is less, even though you are moving the weight twice as fast because the amount of force used to move each load is less- only 20lbs. At 11 rpm the amount of torque is too low to make any more power. of course, cams might help the engine breath better, and therefore make the point at which we see 100% Ev higher in the rpm range, and our engine might still make power at 11 rpm, but unfortunately it doesn't matter because the faster rotation exerted too much inertial forces on the conrod and it exploded. Next time we'll use titanium rods in the super liter.
I hope this helps clear things up and isn't too simple. The thing to see in this exaple is that the push on the piston isn't the same, and will rise and fall along the rpm. the point at which the engine inhales the most air/fuel is when it will have the most push on the piston, and make the most torque, and therefore the most load on the piston.
Will

tig488

iTrader: (18)

very good explanation.

JoneZZZ

iTrader: (1)

Wow....fantastic write-up.

Quamen

iTrader: (14)

350z dyno graph

Sharif@Forged

iTrader: (92)

So Resolute, we actually agree on this....where was my misconception?

An engine that generates 400ft/lbs of trq at 3000rpm is under more load/stress than the same engine making 400ft/lbs at 5000rpm? Correct?

prescience

Resolute, you have no doubt answered my question and I must apologise to all concerned that it is just after 12:00 here in the UK - so I cannot appreciate your efforts

And I have more 'fun' to enjoy.

So sorry and to all that have helped me this year. A Happy 2006

Resolute

iTrader: (2)

No, you have it backwards. You were absolutely right in your knowledge that the peak torque occurs when the engine intakes the max amount of air/fuel in the rpm range, but the msconception is that the rpm has an effect on the compressive loads. It doesn't. In your own example, the explosion that makes 400lb/ft at 3000rpm, will have the same energy in the explosion that generates 400lb/ft in the 5000rpm engine. How do we know this? because both engines generate a FORCE of 400lb/ft of TORQUE on the piston. The exact same. The diference in rpm doesn't make the bang any bigger or less- the amount of air/fuel and needed to produce a 400lb/ft thrust on the piston is the same anywhere in the rpm range on either engine since its the same engine. The size of the "bang" is the same, they just occur at different speeds. The reason that the engine in the original example is going to have more stress is because it must rev higher to make the same power as the other engine, 6100 vs 4300, and the inertial loads at the higher rpm make it more stressed. The max compression loads created from combustion are the same.
Will
EDIT, I find my ability to spell correctly decreases at a linear rate proportional to my alcohol intake. Happy New Years everybody!!

Quamen

iTrader: (14)

look at the dyno graph I posted. This would be a prime instance in which I would wonder about stresses.

Lets change the situation a bit now. Given the dyno graph of which i have provided above, which engine would have more stress at the peak torque.

Situation
1) typical FI VQ35 engine producing around the same torque numbers at whatever the average peak rpm is for a FI VQ35
2) The Fi VQ35 engine dyno graph i have provided

I am leaning towards the fact that the powr levels seen in the dyno graph of that motor are too high of stress levels for the stock internals. My reasoning is that the torque is extremely high at a low rpm. And given the A/F ratio (10.0 or less) the cylinder pressures must be imense to make such high power levels at that given rpm. This was also an engine at about 7-7.5psi of boost. So far i have seen no other VQ35 engines making that much torque at that low of an rpm at that boost with that rich of an A/F ratio. This leads me to believe that there is lots of stress in the engine because the power drops off at that point because the timing is being retarded to much and losing power. If the timing was adjusted correctly, I would imagine torque numbers in the mid 400 range. Would this be to much for the stock internals?

Sharif@Forged

iTrader: (92)

Resolute, what is your educational or profesional background? Is your background in engineering or physics?

overZealous1

iTrader: (27)

you did not completely read or understand my post then. instead of drifting so far away from the original thread, i understood and answered his idea. and yes it is 100% true when dealing with a set amount of air/fuel mixture being introduced to the motor. which will only happen with an N20 INJECTION. meaning, his point holds water only in this condition though. it has nothing to do with rpm or torque peaks or anything. ok, to paint the picture for ya, you inject 300hp of N20 at 1500 rpm, and then inject it at 4500rpm, which one will induce more pressure or wear on components? again, his idea only holds true for a motor that has the air/fuel induced in a way that has nothing to do with cam shaft, or breathing of the motor. yes, the motor will see higher cylinder pressures at lower rpm because the valves are open longer (again this is a time based function) and a set amount of air /fuel being induced.

if you still believe i am wrong, i must not be explaning it in a way you are understanding. this is what he was referring to in his original post. read again carefully if you don't understand what i'm saying.

Resolute

iTrader: (2)

OK, so when is he asking about injecting nitrous?
He asked which engine has more stress, the one making power at 4300rpm or the one making power at 6100 rpm.
Both engines are turbocharged, and everything is identical between them.
I didn't drift from this question.
The answer is the engine making power at 6100. The fact that one makes power earlier does not mean it induces more stress on the engine, because the amount of torque applied to the piston, conrod, and crank is the exact same for each engine according to his parameters. Both engines have 400lb/ft of max torque applied to the engine, and therefore both engines suffer the same exact compression loading. The only difference in load, and therefore stress, is the rpm, and since one engine -according to his parameters- must rev higher to make the same power, then that higher revving engine is the one under more stress.
Now, you are stating that because the valves are open longer at lower rpm, then more air/fuel is going into the combustion chamber. And as I posted earlier, this is grossly wrong. Even with forced induction, this is wrong, and since his example is a turbocharged VQ, this certainly seemed relevant to me, and not drifting from his question.
Now, I think I smell what you're stepping in. With your nitrous example, if you directly inject the nitrous, a "300hp" shot, directly at the intake valve, then the longer the valve is open the more of your shot will make it into the combustion chamber for that one cycle. This holds water. Of course, it has nothing to do with his question. For any turbocharged application, the rpm at which Volumetric Efficiency (Ev), and thereby torque, is greatest is a combination of the engine's own breathing characteristics and the efficiency characteristics of the forced induction method combined. I'll be more than happy to discuss the reasons for this later.
As far as the dyno Quaman posted, the answer to your first question is the same. Take any "typical" FI VQ engine, producing the same torque as yours, and whichever one produces their torque at the highest rpm is the one under more stress. As far as your tuning and whether it is going to grenade, guys like Sharif will be better suited to answer if more than 400lb/ft of torque is too much for the stock engine considering the direct experience they have. I personally think your tuning is an issue because your air/fuel is aweful, but look at the bright side, keeping it rich keeps your temps down and guards against detonation.
Will