Rebuilding ford 302 HO any idleas ?!?! :)

[haulnazz15]

Re: Rebuilding ford 302 HO any idleas ?!?!

Here's what I have in the 351W I just built:

Bored .030
Comp Cams XE256-grind cam
Pertonix Billet Marine Distributor with Ignitor II module and Flamethrower Coil
Edelbrock Performer Intake
Edelbrock 1409 Carb
Osco Center-Riser conversion manifolds

I'd estimate somewhere between 260-280HP. Edelbrock carb couldn't have been easier to setup.

 

[Aloysius]

Re: Rebuilding ford 302 HO any idleas ?!?!

Ohio points out a critical factor in performance..dialing in the ignition timing! All too often this simple but critical component is overlooked. All the trick hard parts won't mean a thing is the timing is only 30 degrees advanced.

This applies to a STOCK engine also. Money is better spent dialing the existing ignition and carb. I personally picked up 5 mph top speed (30 to 35 mph) and increased fuel mileage almost 50% by these simple details.

 

[45Auto]

Re: Rebuilding ford 302 HO any idleas ?!?!

torque is a measurement of force. work equals the integral of F.Ds. By putting weight on the end of a breaker bar, you've done no work unless you move something.

If you had said "force" instead of "weight" it would have demonstrated a better understanding of the basic concepts.

Torque is a measurement of one type of Force just like Power is a measurement of the rate of doing Work.

Per your example with the force being described as a weight, then

Weight = Mass x Acceleration and Torque = Weight x Distance. So

Torque = (Mass x Acceleration) x Distance

No way to determine torque without knowing all three components that make it up.

There are two types of units, base and derived units. Base units are the simple measurements for time, length, mass, temperature, amount of substance, electric current and light intensity. Derived units are made up of base units, for example, position is typically described in units of length. Speed is described as length per time (feet/sec). Acceleration is described as length per time per time (feet per second per second, or feet per second squared). That's why speed is the first derivative of position with respect to time (V = dx/dt) and acceleration is the second derivative of position with respect to time (A = dx/dt/dt), and the first derivative of speed with respect to time (A = dv/dt).

Notice that torque is not a base unit but is derived (calculated) from length and force. In the english system the unit of force is the pound, which itself is derived from mass and acceleration.

At least you're right about the work.

Torque is what is measured on a dynomometer, and it is indeed an actual measurement for engines..it is the ONLY aspect measured along these lines. From the measured torque, horsepower is calculated.[/

I?m not sure what your hang up on measured and calculated is. If your unit is not length, mass, or time then it's calculated. How do you get "feet x pounds" without doing a calculation?? Torque and horsepower are both calculated units. How exactly do you "measure" your torque? Do you actually think all dynos have a moment arm exactly one foot long? Could it be that they really measure some force, and knowing the distance over which it acts the torque delivered by the engine is calculated?

There are lots of different ways to couple an engine to a dyno and provide the load that reacts the force from the engine. But it all comes down to force x distance.

Let's say you're using a dyno with a roller that is driven by the engine flywheel. It's a typical GM engine with a 14" flywheel that's turning at 5000 RPM. You have a 4 foot diameter load roller and your load system (could be water, a brake, magnetic, etc) is putting a load of 600 pounds on the edge of the roller. This means that the engine is also putting 600 pounds of force on the edge of the roller in the opposite direction. The torque required to hold the engine at 5000 RPM is 600 lb x 2 feet = 1200 ft-lb of torque. Does this mean our engine is putting out 1200 ft-lb of torque to you?

Now let's say we break our roller and all we have to replace it is an 8 foot diameter roller. We run the engine again which hasn't changed. The engine is still turning 4500 RPM and putting 600 pounds of force on the edge of the roller. So our load system also still has to put 600 lb of force into the edge of the roller. That's 600 pounds x 4 feet = 2400 ft-lb of torque!! Do you really think we doubled the torque output of the engine just by changing the diameter of the dyno roller?

In the real world, every dyno will report that the engine above is delivering 350 ft-lb of torque. Something (computer) or somebody is converting force x distance to ft-lb. This is because the convention is a one foot lever arm. The unit is ft-lb, not 2ft-lb or 4ft-lb. So regardless of the size of the roller, somewhere it is calculated and normalized to 1 foot.

With the 4 foot diameter roller, the engine torque normalized to a one foot lever arm (ft-lb, remember?) would be:

engine torque (ft-lb) = roller torque (ft-lb) / roller radius (ft) x flywheel radius (ft)

= (1200) / (2) x (7/12) = 350 ft-lb

With the 8 foot diameter roller the engine torque normalized to a one foot lever arm (ft-lb) would be:

= (2400) / (4) x (7/12) = 350 ft-lb.

In the real world, we ignore all the roller torque shenanigans above and just calculate the engine torque directly from the brake load. If our engine is putting 600 pounds into the edge of the flywheel, the brake load (if it's also acting on the edge) will always be 600 pounds. The torque delivered to the dyno will depend on the size of the input roller, but it's irrelevent. Nobody cares how much torque is really going into the dyno, all we care about is the normalized (ft-lb) torque that the engine is delivering.

We know that the edge of our 14" diameter flywheel is delivering 600 pounds of force because of the brake load we're reading. Since the flywheel is 14" in diameter, this means that the engine torque is

engine torque = force (lb) x distance (ft) = 600 x (7/12) = 350 ft-lb.

Since we calculated that our engine delivers 350 lb of force at 1 foot (although it was really delivering 600 lb at 7 inches), and it is turning 4500 RPM, then every dyno will calculate that the engine is delivering 300 HP no matter what the real torque is that?s being delivered to the dyno.

If you now understand how a dyno works you'll realize that the engine never really delivered 350 pounds of force at 1 foot. In the 4' case it was delivering 600 pounds of force at 2 feet for 1200 ft-lb on the dyno, and the 8' case it was delivering 600 pounds of force at 4 feet for 2400 ft-lb on the dyno. If the engine flywheel diameter was 14" like a typical GM flywheel, then the engine was delivering 600 pounds of force at the edge of the flywheel.

To go back to your earlier example, do you really think that putting a "weight" on the end of a breaker bar always provides the same torque?

Let's say we're going to the moon, and we have to turn a fuel valve on our spaceship engine to launch from earth and again to launch from the moon. It takes 30 ft-lb of torque to turn, so we cleverly design a mechanism that turns it by hanging a 32 pound weight on a 1 foot bar. How long will it take us to return from the moon?

The answer is that you never will. Your fuel valve won't turn on the moon.

Simply put, the problem is "Calculate the torque (ft-lb) exerted on earth by a 32 lb weight hanging from a 1 foot bar. Calculate the torque (ft-lb) of the same set-up on the moon."

We'll agree to use the NASA gravity table which rounds acceleration due to gravity on Earth to 32 ft/s/s, and on the moon to 5.3/ft/s/s.

http://settlement.arc.nasa.gov/teacher/lessons/bryan/microgravity/gravback.html

On earth, you would have 32 x 1 = 32 ft-lb of torque. The valve would turn, you'd be on your way. You calculated the torque correctly.

That's because the english unit of mass is the slug. You placed one slug of mass on your lever, which weighs 32 pounds on earth. Weight is a force which depends on gravity. A mass of one slug exerts 32 lbs of force on earth. 1 slug x 32 ft/s/s = 32 lb.

On the moon, the force will only be 1 slug x 5.3 ft/s/s = 5.3 lb.

In other words, things only weigh 1/6 as much on the moon as on earth.

So the torque your mechanism provides on the moon would be 1 slug x 5.3 ft/s/s = 5.3 ft-lb. The valve requires 30 ft-lb to turn so you're stuck on the moon. You didn't calculate the torque provided by your 1 slug mass correctly.

It would be even worse in orbit (free fall). Need to loosen a bolt on the space shuttle that needs 30 ft-lb of torque? Hang the space station from a 1 foot breaker bar attached to your bolt. The space station's weight on earth would be a little over 750,000 pounds. But hanging from a 1 foot (or 10 foot or 100 ft etc, etc) breaker bar it won't turn that 30 ft-lb bolt for you. We can see why if we calculate the torque:

Torque = Force x Distance = (0) x (1 foot) = 0

The space station and space shuttle are both in free fall, so the force relative to the shuttle is zero.

Hopefully you're beginning to realize that knowing that torque is a calculated number and knowing how it's calculated could be important?

You cannot get "horsepower" without torque, and torque is what is actually measured.

People have been getting horsepower without torque for millions of years. See above for discussion about "measuring torque" (repeating your view multiple times won't make it right - perhaps for discussion you could provide an example of how you think it is "measured" if you don't agree with mine?). You don't even need a horse. "Horsepower" is simply a convienient measurement of power which equals 33,000 lb per foot per minute. You may want to review the history of James Watt and how the term "horsepower" was derived. A team of horses lifting 33,000 pounds one foot in one minute is one horsepower. A thousand Egyptians dragging a 33,000 pound block of stone up the side of a pyramid one foot in one minute is one horsepower. The same thousand Egyptians dragging a 132,000 pound block of stone up the side of a pyramid one inch in 48 minutes is one horsepower.A rocket lifting 33,000 pounds 1 foot in 1 minute is one horsepower. No torque anywhere in sight. As long as you're measuring lb-ft/min all you have to do is divide by 33,000 to get horsepower.

If you want "horsepower" at 4500 rpms, you gotta have the necessary TORQUE at 4500 rpms to produce the horsepower you want.

That's true. It's also true that if I have an ungodly amount of torque and want a given amount of power, I gotta have the necessary RPM to produce the horsepower I want. A 500 lb weight hanging off a 2 foot breaker bar connected to your prop (1000 ft-lb) isn't going to move you very fast.

Power is the key to performance. Torque and RPM are both irrelevant except as you use them to provide power. Either one will work, whichever provides the most power will be fastest.

If you think torque is the secret, try running your 4500 RPM motor against a gas turbine powered boat some day. Your 4500 RPM motor may have 350 ft-lb of torque compared to the gas turbine?s maybe 5 ft-lb at 4500 RPM (they actually won?t even idle that slow).

Geared 1.5:1 means you?re putting 525 ft-lb to the prop, and the prop is turning 3000 RPM. The gas turbine will be making over 1800 HP at 60,000 RPM. That?s only 158 ft-lb of torque. But it?ll be geared 20:1, so that the motor is turning 60000 RPM when the prop is turning 3000 RPM. It?ll be putting 158 x 20 = 3150 ft-lb of torque into the prop. Which do you think will win?

There?s really no need to go through all the torque calculations. You?re wasting a lot of time. In the example, the 4500 RPM, 350 ft-lb motor put 525 ft-lb to the prop at 3000 RPM. We can calculate the power the motor put out:

HP = 350 x 4500 / 5252 = 299.89 HP

Which will be exactly what went into the prop (we?re ignoring drive train losses)

HP = 525 x 3000 / 5252 = 299.88 HP

You?ll remember that the torque our turbine put into the prop at 3000 RPM was 3150 ft lb. So at 3000 prop RPM, our turbine had 3150/525 = 6 times the torque. Since power = torque x RPM / 5252, it was obviously putting 6 times the power into the prop since the prop was turning the same RPM.

If we were clever, we would have noticed that our original motor made 300 HP. We would also have noticed that the turbine made 1800 HP. Notice how 1800/300 = 6? Look at the paragraph above and notice the torque ratio at the prop. It was also 6! What a coincidence!

The bottom line is that I didn?t need to know RPM or torque or any of this stuff to tell you the relative performance of these motors. All I needed was the horsepower numbers. The turbine has 6 times the power, so no matter what you do, it?ll deliver 6 times the speed of the other motor.

If you want to know acceleration, you?ll need a power curve showing the engine power (as geared) at each speed you?re interested in. Say 5, 10, 15, ?.. 70, 75, 80 MPH etc. The engine with the most power at that particular speed will accelerate faster.

There?s no difference between an engine with 350 ft-lb of torque at 4500 RPM, one with 700 ft-lb of torque at 2250 RPM, and one with 17.5 ft-lb of torque at 90000 RPM. All three have exactly the same amount of power and will do exactly the same amount of work. Just means that the appropriate gears will be required depending on what you desire your output RPM to be.

 

[45Auto]

Re: Rebuilding ford 302 HO any idleas ?!?!

Ever wonder why the numerical equivalent horsepower and torque curves cross at 5252 rpms? It's FROM the FORMULA.

I?ve never wondered because I?ve always known as far back as I can remember. It's a totally arbitrary number. They cross at 5,252 because the distance around a circle is 2 x PI x R, and James Watt decided that 1 foot was a good number for a moment arm, and Watt decided that a horse could lift 550 pounds one foot in one second in the late 1700?s. 550 pounds one foot in one second is the same as 33,000 pounds one foot in one minute (that's one horsepower).

In a nutsack, (2 x PI x 1) / (550 x 60) = 6.28 / 33,000 = 1/5,252. If he?d decided that 6 inches was a good moment arm they would cross at 10,504. If he'd decided 2 feet was a good moment arm they'd cross at 2,626. If he'd decided that a horse could only lift 540 pounds one foot in one second they'd cross at 5,159. If he'd have lucked into a strong horse that could lift 570 pounds one foot in one second they would cross at 5,446. I very seriously doubt if you have any more understanding of it other than "It's FROM the FORMULA" or you would realize how silly some of your statements are.

James Watt decided a few hundred years ago to normalize his units for power to 1 foot and 1 minute for his "horsepower" thing. If you?re talking rotational velocity and standardizing on a 1 foot lever arm, then your distance becomes the distance around a circle with a radius of 1 foot. Circumference of a circle is 2 x PI x R, so the standardized circumference is 2 x PI x 1 = 6.28. Didn't have to do anything about the minutes because the standard unit of measurement for rotation was already revolutions per minute.

Basic power unit (not HP yet) = ft-lb/sec, so to use the power of a 4,500 RPM, 350 ft-lb engine in an example we have to multiply by 60 to get it into the basic power units of ft-lb/sec. 4,500 x 350 x 60 = 94,500,000 ft-lb/sec

Remember, James Watt standardized HP to a one foot moment arm a few hundred years ago. To convert from basic power units to HP first we have to multiply by 6.28:

94,500,000 x 6.28 = 593,460,000 ft-lb/sec

Watt also decided that one horsepower = 33,000 ft-lb/min. So to get to minutes we'll have to divide the power by 60:

593,460,000 ft-lb/sec / 60 sec/min = 9,891,000 ft-lb/min

Then to get to HP, we have to divide by 33,000 ft-lb/min:

9,891,000 ft-lb/min / 33,000 ft-lb/min = 300 HP.

So the entire equation is:

HP = torque x rpm x (6.28 / 33,000)

Notice that the division by 60 to convert to minutes that we did a couple of steps above is already contained in the "revolutions/min" term since 60 revolutions / minute = 1 revolution / sec.

If you did good in math, you?ll notice that the 6.28 and 33,000 are what are called ?constants?. In other words, they never change. So why multiply by 6.28 and divide by 33,000 every time we want HP? Do it once and be done. 6.28/33,000 = 1/5252. So our equation is now:

HP = torque x RPM x (1/5252) which is the same as

HP = torque x RPM / 5252

We can check it with our 4500 RPM, 350 ft-lb motor:

HP = 350 x 4500 / 5252 = 300 HP

Another way to put it is that if your buddy tells you that the power of his engine is 94,500,000 ft-lb/sec, he's really telling you that he has a 300 HP engine ....

Ain't math cool?

Any more questions? Can you tell that I was bored for the last hour waiting to go home?

 

[Lyle29464]

Re: Rebuilding ford 302 HO any idleas ?!?!

Your outdrive will not take a lot of pain. I would stick with a truck,marine or RV cam. If your engine requires a high idle speed and you shift at 900+ rpm
or you are docking at 10 knots you will not be happy. I would really think about the drive when I bought a cam and each time I picked up a bit of speed.

 

[Aloysius]

Re: Rebuilding ford 302 HO any idleas ?!?!

Holy crap..talk about BORED! Let's see..I took statics and dynamics when plate tectonics was just being recognized. I'm just trying to keep it simple, for the everyday guy. Few people realize the relationship between torque and horsepower, and the effect rpms has on both.

No argument on the technicalities and details. Thankfully you didn't drag the thermodynamic side into this..........

45auto..your assignment for tomorrow is to explain the relationship between energy in (fuel), thermal efficiency, internal friction, compression ratio, and..and..(and anything else I can think of to obfuscate the project).

Just kidding. But it IS important for the average gear head to realize the relationship between "torque" and "horsepower" as it relates to the internal combustion engine.

Ahh..college in the '60's. No wonder everyone smoked a big fattie before going to "electrical and magnetic field THEORY".

 

[45Auto]

Re: Rebuilding ford 302 HO any idleas ?!?!

I took statics and dynamics when plate tectonics was just being recognized.

Maybe if you tried again now, you could pass them!

 

[chaparall villain 2]

Re: Rebuilding ford 302 HO any idleas ?!?!

ok i am gonna give you what i consider helpful info .. i never realized how complicated it can all be until i rented a shop next to a machine shop with a mechanical engineer that owned and operated it .. alot of stuff sounds like its great for horsepower or torque but if you buy a bunch of stuff that doesnt work together you end up with a bunch of problems and extra expense ... find a good machine shop that has a name you can trust and go talk to them about what your trying to do and see what suggestions they make ... i see people come in next door with all these parts they bought online and all of a sudden the valve springs wont work with the cam or the lift is too much for the heads and they need more head work ... or the pistons they got make the compression too high and unable to run on pump gas . could be worth a discussion with someone who figures all these things out before you go out and spend a bunch for nothing ... not tryin to bring you down but i know on my engine i learned so much from my machine shop that i would have never known otherwise .... good luck

 

[Jaysad]

Re: Rebuilding ford 302 HO any idleas ?!?!

Well, I have to thank you all for the info.!
I think I have alot more planing to do before I get going on this.
Wow, that torq. & hp thing was pretty cool too! Thanks

I'm going to put a 351w remaned long block. Most of the ones I've look into are bored 20 -30 over. I'm adding a alum. holley intake manifold and a 650 holley 4bbl. carb. I think this is a good start for now.
I'm going to keep this info. so I can refur to it later, again thanks to all!