Compression and boost?

Im not a pro engine builder but, the research I have done on forums, dyno shops, & machine shops says that a stock 302 block can handle about 540 hp. As for compression ratio, you might not be able to run as much boost as a 8 or 9:1 motor but, you could run an inner cooler, meth injection, or both (which is what I will be doing). My compression ratio is 9:1 and I have a 76 mm rear mounted turbo that I hope to make about 500 hp at the crank once I get it all back together. I might very well find out my research is wrong but, thats my plan. hope it helps.

How much horsepower are you looking for?

450 hp to the wheels + about 15% drive train = 517 at the crank. That should be within the limits of what a factory 5.0 block can handle from the info I have found anyway. I had a formula on my old laptop that could calculate how much boost was needed to make X hp at Y rpm but, I dont have that laptop anymore. If my buddy still has the formula I will post it for ya.

450rwhp with 15% driveline loss is 529bhp. Divide 450 by 0.85 instead of adding 15% to 450.

As far as how much boost will increase power by what amount, 15psi will roughly double horsepower with good intercooling. 7.5psi will make 50% more, 3.75psi will be roughly 25%, etc....about 6.8% per psi.

3ninety2 said:

Hey, I need help on deciding on whether or not to put a blower on my new motor. Just finished a 331 Stroker w/ Edlebrock Performer heads and the compression is about 10 : 1. 9000 series Scat crank, forged pistons & rods. Block is a Summit 347 block, looks to be a mexican block. Also running a main cap girdle. A local shop in Atlanta, Ga. told me a 302 can only handle 400+ hp before the main webbing gives way. I,ve been running 351 windsor motors for almost ten years now with no problem. Any advice would be appreciated.

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I highly, highly doubt you have a Summit block that is "hecho en mexico." Also, 351W blocks are stronger than 5.0s anyway. A "347 block" is just a 302/5.0 block with a 0.030" overbore. The stroker determines whether it is a 306, 331, or 347.

With 10:1 compression already, I would just get better heads and go NA.

65ShelbyClone, thanks for clearing up my math mistake. I see how I went wrong with it. 450 was already the answer to 15% loss of 529 so, By taking 15% of 450, my answer was lower 517. I dont like to put out bad info. sorry about that.

You could always save that vortec for the 427. what a beast that could be. I would like to go with the Dart 363 short block myself. That way, all my 302 stuff will fit. Its supposed to be able to handle 1000 HP. My turbo can support upto 800 HP. good luck with it which ever way you go.

You should be fine at 450 rwhp on a 302. I made over 420 myself with a turbo 302. However, the 85% drivetrain loss isn't a good number to use because the supercharger also take hp to run. So, you're actually going to be pushing the block harder than a nitrous, n/a, or turbo engine would at the same power level. Exactly how much is hard to say and depends on the efficiency of the supercharger and how much boost it has to push. However, I think 450rwhp is still rather conservative.

You need to start thinking about splitting the block when you hit 500rwhp. And have a replacement fund ready if your at or over 600rwhp.

Chris

10:1 is a bit high for a supercharged application on pump gas. I'd stay away from a combination like that unless you plan to run race gas or e85.

Well, it is size of bore AND stroke that give you Cubic Inches, not one or the other...it's a combination of both.

If you think you have Mexican Block, which coming from Summit, I doubt it, all youhave to do is look at lifter valley. It will tell you. If Mexico, then will say made in mexico. If missing that, then simply a 302 block, and probaly a late model roller block at that.


Not to start up the "parasitic loss" thread, but I have never been one to subscribe to a 15% or 20% loss theory.
My argument has always been this:

If you take a 500 at fly, say running through a Turbo 400, C4, TKO, whatever put on chassis dyno, and you subscribe to the say 15% number...then losing 75 Horsepower, right?
Now, take same car, and you have dropped in a 1500 HP motor, and you subscribe to that 15% theory. So, you mean to tell me that the SAME TRANSMISSION AND REAR END is now accounting for 175 HP loss?
To me, it takes a certain amount of power to drive a tranny/rear end, and saying it always a % of the number you make just doesn't add up.


Anyways, back to OP....10:1 with boost is possible, just going to be hard to get away with just pump gas without using meth or the like. Typically the 10:1 crowd and boost is reserved for race gas and teh track. On the street, going to be hard to keep together, especially come summer time and the heat...food for thought.
Personally, if it was me...would either lower compression, or keep NA...

You can work it out with 10:1, but I agree that it's far from ideal. Low boost, say 6psi, and conservative timing and you'll be fine.

Gen414, I agree with what you're saying. It's probably not a valid assumption when you start talking about big power cars. It's certainly not accurate when you're talking about an auto with a healthy stall.

Gen414 said:

1.) Well, it is size of bore AND stroke that give you Cubic Inches, not one or the other...it's a combination of both.

2.) If you think you have Mexican Block, which coming from Summit, I doubt it, all youhave to do is look at lifter valley. It will tell you. If Mexico, then will say made in mexico. If missing that, then simply a 302 block, and probaly a late model roller block at that.

3.) Not to start up the "parasitic loss" thread, but I have never been one to subscribe to a 15% or 20% loss theory.
My argument has always been this:

If you take a 500 at fly, say running through a Turbo 400, C4, TKO, whatever put on chassis dyno, and you subscribe to the say 15% number...then losing 75 Horsepower, right?
Now, take same car, and you have dropped in a 1500 HP motor, and you subscribe to that 15% theory. So, you mean to tell me that the SAME TRANSMISSION AND REAR END is now accounting for 175 HP loss?
To me, it takes a certain amount of power to drive a tranny/rear end, and saying it always a % of the number you make just doesn't add up.

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1.) Read this again:

65ShelbyClone said:

A "347 block" is just a 302/5.0 block with a 0.030" overbore. The stroke determines whether it is a 306, 331, or 347.

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2.) Mexican blocks also have a raised boss on the front above the oil filter. They actually have one on each side, but I think that one is more obvious.

3.) It does add up as Mustang5L5 said, although it is not limited strictly to inertia. The more power being transferred through a drivetrain, the higher the contact forces and friction are as well. That's why driveline loss is always a percentage of engine output and not some standard number applied to a specific drivetrain.

65ShelbyClone said:

1.) Read this again:



2.) Mexican blocks also have a raised boss on the front above the oil filter. They actually have one on each side, but I think that one is more obvious.

3.) It does add up as Mustang5L5 said, although it is not limited strictly to inertia. The more power being transferred through a drivetrain, the higher the contact forces and friction are as well. That's why driveline loss is always a percentage of engine output and not some standard number applied to a specific drivetrain.

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1)What am I reading again? I said bore and stroke equal cubic inches. So????????

2)Lifter valley is easiest to read/look for/at

3)Well, not a scientist, but I don't see it. Not saying Im right, but, I don't feel Im wrong either..

Gen414 said:

3)Well, not a scientist, but I don't see it. Not saying Im right, but, I don't feel Im wrong either..

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It's just like calculating rolling resistance for a simple wheel and surface. Provided the wheel and surface do not change, as the weight load on the wheel increases, the force required to keep the wheel rolling at a constant speed (or initiate rolling from a standstill, ignoring inertia) increases proportionally. It never gets to a point where rolling resistance stops increasing even if load increases further.

So yes, as engine power production increases, losses through the drivetrain increase proportionally. The "% method" isn't perfect, but it's the closest way to guess given most any measurable power level.

Well, I have to say that it is my feeling that for a given MANUAL trannied car you will not have an accurate estimation of rwhp for a 1500 hp car if you first determine the % drivetrain loss for the vehicle at say 300hp.

What is drivetrain loss? I can think of four things that contribute, though there are probably more:

Well, a torque applied at one end of a static object will apply an identical torque on the other end. However, we're not dealing with static objects, but accelerating objects. The acceleration in and of itself "consumes" some of that force meaning that the force applied at the other end of the object will now be less the torque required to accelerate that object's mass into rotational velocity. I think it's somewhat difficult to determine the force required because while we could estimate the rotational inertia of a given drivetrain, the rest of the equation would depend on the rate of acceleration of each component of the drivetrain, each of which would depend on the rate of acceleration of said vehicle. They would each be proportional and dependant on the gear ratios/tire radius involved. This would, therefore, not necessarily be a constant percentage of the power applied, but would depend on where you are in the powerband, which gear you're in, and . As a general rule, more force equals more acceleration, so this requirement as Mike mentioned should increase with the drivetrains rate of acceleration. Inertial dynos, therefore, would show an increase in measured torque loss that increases with the amount of torque produced. However, I'm uncertain whether or not this could accurately be represented with a constant percentage of the rate of acceleration. On a dyno that held a constant velocity, however, the loss of force due to this phenomenon would be 0.

Another loss in the drivetrain would be the loss of torque due to friction between the gears in the transmission, and the friction between the ring/pinion gears in the rear-end housing. This too, would increase proportionally with the amount of torque applied, and would be accurately represented by using a constant percentage of that torque. That's because the kinetic friction coefficient is that constant, and the only other information required to determine the friction involved is the normal force which would be proportional to the torque applied to the system, and which would account for the angles of the gears and the gear ratios used as gear ratios directly multiply the normal force.

There would be a some amount of loss in the hysterisis due to deflection in the tire, too. This loss of force is the reason that some tires allow a car to get better gas mileage than the previous tires did. With most street tires, the force loss should not be noticeably different when more force is applied. The biggest factors in tire deflection include the weight of the vehicle on the tire, and the speed of the tire. Neither of these need change in a higher HP application. However, non-standard tires like bias ply tires used for racing will deflect significantly due to the torque applied and as a result the force loss here should increase as the torque applied to the tires does. However, I would expect this loss to be relatively small by comparison to the other factors already mentioned because the tire only deflects once as the power is applied and then begins to recover its normal shape slowly as torque decreases through the powerband, and then more so when the driver gets out of the throttle. However, this is really only one long event for each molecule whereas for the other factors discussed (speed and weight) each rotation of the tire causes a deflection and recovery.

The last loss of force that occurs to me is the force lost in the friction of the bearings of the moving componenets in the drivetrain. Though it's a relatively insignificant force-loss overall, some of it will also increase with the force applied, but some of it will not. That's because the force applied also adds weight to some of these bearings like the ones in the transmission, u-joint, and rear-end. As load is applied to the gears, they in turn add load to the shaft that they are on, which adds weight to the rollers on their bearings. Other bearings do not take the added weight because the force is not applied directly to them, only a rotational force. When that happens, no extra weight/load is applied to the bearing, and only the twisting force (torque) is applied to the component itself.

I'm sure this is getting to be a bit of a novel now, but it's interesting to me because I believe one side of the argument, though I've never actually thought through the forces and losses involved. This is not an argument so much as an open discussion. Feel free to add any input or correct my observation.

NikwoaC said:

It's just like calculating rolling resistance for a simple wheel and surface. Provided the wheel and surface do not change, as the weight load on the wheel increases, the force required to keep the wheel rolling at a constant speed (or initiate rolling from a standstill, ignoring inertia) increases proportionally. It never gets to a point where rolling resistance stops increasing even if load increases further.

So yes, as engine power production increases, losses through the drivetrain increase proportionally. The "% method" isn't perfect, but it's the closest way to guess given most any measurable power level.

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Yeah, I can see your point..still not sold

However, I DO realize I am the minority on this, and the majority goes by that, so obviously on to something. Call me hard headed...