Question re raising compression '95 f355

I've actually looked into this a while back when researching 308 valve springs. That is not the case, the energy returned during valve closing is nowhere near that which is required to open. Crane cams (and I'm sure others) have a specific valvetrain dyno which they measure load required to open valves for Cup engine valvetrain testing and simulations. As I understand it is a pretty trick setup as it spins a lubricated cam attached to an entire valvetrain which allows close monitoring of components as a whole, and torque required to spin the cam is how they measure the parasitic drag from a particular valvetrain. The general rule is to use only 5% more valve spring nm than that is what is required to control the valve.

 

Da ya think?

 

It is actually torque which accelerates a car. HP is what is required to keep a car moving at a constant velocity to over come the drag forces. However, torque and HP are directly related.

HP = T x RPM x some constants.

The factor of RPM is an indication of why the HP and T curves vs RPM are important. A car with high low end torque accelerates rapidly with the engine at low RPM, and will also have higher HP at low RPM. Note also that an engine which has constant torque has HP that increases linearly with RPM.

 

I said the spring returns a -portion- of energy spent, which is correct. I did not state there is zero net loss.

And you definitely want more than 5% overhead on the spring. (lose control at 100lbs on the seat so install at 105?).

You don't want to be right on the edge of valve control as the springs will lose some strength over time.

Point is we are dealing with a factory multi valve head with lightweight components capable of 8500+rpm with no loss of valve control. There simply isn't 40-50hp to be found by losing a few grams per valve and a few psi of spring pressure.

 

John, you can have all the torque in the world but if you have no rpm,the car won't accelerate quickly. I can illustrate mathematically, but the gist is that since we have gear reduction, hp indicates the potential for torque at the rear wheels. Two engines making the same hp have the same potential for rear wheel torque regardless of what the crank torque is (when I say rear wheel torque, I mean crank torque after it is multiplied through the transmission and differential, not crank torque minus drive train loss).

 

If that were true it would have been done and standard procedure by now.
These motors have little left on the table in few areas. Nip and tuck for sure

Weight loss overall is the best performance on these cars

 

Actually what you said was that the springs are not a high source of drag. But when you look at the entire valve train the force required to open the vale is the required to compress the spring and to accelerate the mass of the valve, the tappet and some effective mass of the spring. The only part recoverable is a fraction of the spring force which is greater than that required to close the valve and keep the tappet in contact with the cam lobe. At some RPM the spring force will not be sufficient to do that and RPM there would be no energy recovered. At very low RPM the acceleration of the masses in the valve train is more or less insignificant and almost all the force or energy applied to compress the valve spring can be recovered, but as the RPM rises less and less is recovered.

 

Is it required to quote everything I've previously said in order to continue a conversation?

"Actually what you said" and "that is not the case" would imply I said something to the contrary, which I didn't.

And yes, compared with total drag of an engine, valve springs are not a high percentage. That title belongs to the pistons/rings.

I do understand that returned energy will decrease as rpm increases but that will be some portion unless valve spring overhead is zero. (especially at peak tq/hp rpm)

A 50hp spintron can turn an ENTIRE DOHC V8 10,000+rpm. There is not 40-50hp in valve weight/spring parasitics. Not even close.

 

Agreed!

 

These guys have engine upgrades

Piston - TODA RACING

 

^^ Looks like really nice stuff. Problem is take a normal retail price, multiply that by 10, then multiply that by crazy and you now have TODA pricing

 

Exactly. Sillyness

 

As others have noted, the biggest gains on this car will be in reducing weight (assuming that you do not use forced induction, E85, or race fuels e.g. VP C16) and not from increasing the compression ratio which is already fairly high at 11:1 compression, dramatically increasing RPM, or stuffing monster camshafts in there. I will concede that there is some room to get bigger camshafts in there, but, then you need to look at the whole valve train which is stressed already considering the 8500-8750 RPM redline. Few V-8 engines spin this fast for a reason, it's hard on components and every few RPM more just adds to the problem exponentially.

Reducing weight, by contrast, reduces stress typically unless that weight is taken from some component that needs the mass to provide rigidity or structure. Every unused pound you remove from the car is helping. Especially if balance of the vehicle from to back is equally affected. There are some things that you won't likely pitch, despite the weight savings, for instance the air conditioning (if you live in a hot climate). However, if you live somewhere when it never gets too hot nor too cold, you might consider it. Getting 250 lbs out of the car (a sizable amount and not trivial by any means) would be the same as gaining 35 HP, also not a trivial amount. I will grant you that there are things that can be done on the exhaust and with the engine management that could gain 35 HP, but it would probably cost you more than reducing the weight by 250 lbs.

 

Not exactly correct. Why do you think a car like a Tesla has such great acceleration? Electric motors generate max torque at zero RPM. As far as what can be shown mathematically, it's just Newton's laws. if you want to talk math,

F = Ma > a = F/M

If the torque at the rear wheels is T, the, assuming no wheel spin, F = T/r where r is the distance from the center of the axis to the road surface and

a = T/(r x M)

Now, assuming the rear wheels are turning and the cluck is locked up we could write the torque at the rear wheels as

T = HP/rpm where rpm is the rear wheel RMP, not engine.

and then

a = HP/(rpm x r x M) and we could be lead to believe that more HP means greater acceleration. But then we have to consider a simple example. Take a car with 500 HP at 5000 rpm and a final drive ratio of 2.5. Note that gear ratio has ne effect on HP so the rear wheel HP would be the same as the crankshaft HP (less gear losses). Assume it is accelerating passing through an engine RPM = 5000, wide open throttle. The rear wheel rpm would be 2000, and

a = 500 / (2000 x r x M).

Now change the final drive ratio to 5. Still cruising through 5000 RPM, wide open throttle. The car would be moving slower since the rear wheel rpm would only be 1000. Since changing the gear ratio does not change rear wheel HP the acceleration would be

a = 500 / (1000 x r x M) or 2 time the acceleration of the first case. Same HP, twice the acceleration because it is the torque that accelerates the car.

Now you can go back to the first case and say that you could change the engine to one that develops 1000 HP at 5000 RPM and say the cars acceleration would be

a = 1000 / (2000 x r x M)

and conclude that doubling the HP leads to twice the acceleration. That is true, but it is because the 1000 HP engine develops twice the torque. So yes, you can think in terms of HP, but that's where first principals come in. The more primitive quality is torque. HP is derived from toque, not the other way around. This is born out be the equation

HP = T x RPM which still has meaning when the RPM goes to zero, i.e. finite torque at zero RPM yields zero HP, where as

T = HP/RPM does not. A rotating system has zero HP at zero RPM, but can have finite torque but

T = 0/0 is meaning less.

So, yes, you can write acceleration in terms of HP but it's a manipulation of the fundamental relationship,

F = T/r = Ma.

 

I have to say I'm enjoying this 348/355 forum guys. Lots of good discussion.


I've just gone ahead and ordered a set of valves from Supertech after verifying their valves are indeed a few grams lighter each than stock. Their intake valve features a bit more shaving off the stem just above the valve head which should yield a nominal amount more flow.

I'm still on the fence about bumping the compression a touch because as I look at this thing, with its current displacement it is indeed pretty well on the limit for a street friendly arrangement. What I'm looking for is about 35-50bhp more in the mid range. I'm thinking just starting with a fresh engine with properly built and slightly flowed heads will be a good base to tune the ECU a bit to try and iron out the mid range.

That said, Steve at Topend in Hollywood has already designed me a set of 11.5:1 JE 355 pistons ready to go if I want to pull the trigger....and Andre at Cat Cams and I had a discussion yesterday and he actually has a good Ferrari master (not advertised online) that he suggests would be good with a stock valvetrain. It has a bit more lift (9.9mm total) and only a few more degrees duration above stock but the suggestion is to close the intake valve on the new cam same as the stock cam which would largely maintain dynamic compression. That with a set of Crower rods and I believe this might be the ticket.....

 

The problem with just raising CR is that it a diminishing returns thing. Increasing CR increase thermal efficiency but the higher it gets the smaller the incremental increase become.
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At this point my only reason for slightly raising compression would be to retain stock dynamic compression if I use a cam that closes slighty later than stock cams. It won't take much if I go for a cam that closes only a coule degrees later.

 

You are mis-using gearing in your example. You would never gear two motors with different HP (and torque) peaks the same way, at least not if your goal was maximizing acceleration and top speed.

Let's compare gasoline motors on a level playing field. We'll take two motors, both making 400hp. One is a 7 liter making 400hp at 4700rpm, the other a 3.6 liter making 400hp at 8250rpm. Both are geared to make peak HP at 55mph so that the 1-2 shift can occur at or just past 60mph (which is somewhat typical for sports cars as they want to make that 0-60 run without a shift slowing it down). We'll assume 25" tall tires which would be turning at ~740rpm at 55mph.

So, the 7 liter engine will need total gearing of 4700/740=6.35:1.

And the 3.6 liter engine would need total gearing of 8250/740=11.14:1

That puts them on completely equal footing using maximum mechanical advantage. Gear the 7 liter motor down at all and it doesn't make 60mph in 1st gear, gear the 3.6 liter motor up at all and it will shoot past 60mph still in 1st gear.

Now lets look at the torque at the rear wheels with those two motors.

The 7 liter motor makes 450lbs ft of torque at 4700rpm. Multiply 450 by 6.35:1 and it is putting ~2850lbs ft of torque at the rear wheels.

The 3.6 liter motor makes a puny (by comparison) 255lbs ft of torque. Multiply 255 by 11.14 and it's putting ~2850lbs ft of torque to the rear wheels.

Exactly the same.

Now lets up the power in the 3.6 liter motor to 420hp at 9000rpm. Torque will actually drop 10lbs ft to 245 lbs ft. We re-gear to have the exact same shift point as the other two and make peak power at 55mph. This gives us a gear ratio 9000/740=12.16:1. Multiply 245lbs ft by 12.16 and it's putting 2979lbs ft to the rear wheels.

2978/2850=1.05, so 5% more torque to the rear wheels.

Not coincidentally, 420/400=1.05, 5% more power.

HP tells you exactly how much potential you have for rear wheel torque when geared properly. Because we have a transmission and a goal for a certain speed and not just instantaneous acceleration from zero or a certain engine rpm, the absolute crank torque value does not matter. It's the torque at an rpm that matters, and torque at an rpm is HP.



An electric motor is a completely different discussion, but even so it's the HP that matters in the end. Look at the top speed of the Tesla cars compared to a typical gas motor sports car. It is WAY lower. If you geared a gas motor car with the same HP as the tesla to have the same top speed, it would have roughly the same acceleration. It would just have crazy high gearing and get crappy gas mileage. The beauty of the electric motor is that it's elasticity in terms of making power at both very low and very high rpms, where a gas motor doesn't. A gas motor with a very wide range cvt would perform on-par with a similar HP electric motor, though.

 

Good post Pete!

I always thought a good real world example was 99 F355 vs 99 C5 Corvette. Similar power/weight, much different peak torque/gearing but nearly the same 1/4 mile time and trap speed.

 

Ok, Pete, let's take gearing out of the picture all together. At what RPM will this car accelerate most rapidly, regardless of gear ratio? At 3500 RPM because that is where the max torque is. Acceleration goes like torque. Now you can come back and sat if you increase the HP across the entire power band that a car with this modified engine will accelerate faster than one with my example. True. But that is because the modified engine produces more torque. And the car with the modified engine will accelerate fastest at the RPM where the modified engine is producing the most torque. On the other hand, you would use HP if you wanted to calculate the top speed of a vehicle. For example, considering only aerodynamic drag the drag force on a vehicle would be proportional to

Fd =Cd x V^2

where Cd is the drag coefficient, V is the vehicle velocity and Fd is the resulting drag force. The horse power dissipated by overcoming this drag is

HP = Fd x V / 33000 = Cd x V^3 / 33000

From which the max vehicle velocity can be found.

I haven't mentioned gear ratios or shift points.
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My point was that you CAN'T take gearing out of the picture or gear two engines with different power curves same way. You need to use gearing to maximize force. If you gear two engines to generate maximum force at a certain output rpm, it will always be at the hp peak and the engine with the most hp will generate the most force at that output rpm. Always.

If your target is maximum force at 3500 rpm output speed, you gear the motor to operate at its hp peak while outputting at the desired rpm.

So above, if you use a direct (1:1) output at its torque peak of 3500rpm,you get 410lb ft of torque at 3500rpm. If you gear it to operate at its hp peak while outputting 3500rpm,you will use a 5500/3500=1.57:1 gear reduction. This yields an output of 1.57x335(torque at hp peak)=527lbs ft of torque at 3500rpm. That's 527/410=1.29 or 29% more torque.

Now if we take the peak hp and divide by the hp at torque peak, 355/275, you get, not coincidentally, 1.29, or 29% more horsepower, exactly the force difference above. At any given final output rpm (NOT CRANK),the engine will provide the most force operating at it's hp peak. There is no condition where that is not true. That is why hp is the important number and not torque (without rpm). HP denotes the peak potential for force generation given proper gearing.

If you want to maximize final output torque, you maximize hp. That can be either through increasing torque at the same rpm,or increasing rpm at the same torque, or (ideally) both.

 

CV transmission

 

That would be ideal .

 

CV is just horrid right now, maybe in the future something will plunge it forward

 

Take the curves I posted. I agree that if I wanted to maximize the acceleration at some speed, say 60 mph I would have some over all gear ratio so that at 60 MPH the engine was running at 5500 RPM, where the HP peak is. That is in agreement with what you are saying. If the gear ratio is G1 then the rear wheel torque would be G1 times the crankshaft torque, or about G1 x 340. But that factor, G1, multiplies the torque over the entire RPM range, so at 30 MPH where the engine would be reving at 2750 RPM the torque at the rear wheels would be G1 x 400 and the acceleration would be greater than at 60mph. So yes, acceleration at 60 MPH is maximized by gearing the car so that max HP is applied at 60 MPH, but the car still accelerates faster, with the same gear ratio, at any speed between 30 and 60.

In summary, to obtain max acceleration at a given speed then the engine must be producing max HP at that speed. But that is not necessarily the max acceleration of the car over any given speed range. It would only be the max acceleration if the max torque and HP occurred at the same engine RPM. Regardless of how you gear this car, it will always accelerate fastest, with a given gear ratio, when the engine it running 3500 rpm.