8 cylinder to 4 cylinder in new DC V8's

I guess insults are the best you can offer when you don't have any facts.

Yes. The forumla for a cylindrical shell's volume is: V = 2*pi*r*h*w, where r is the radius of centroid of the revolved rectangle, h is the height of the shell and w is the thickness of the shell. I checked a few places and found typical piston/cylinder clearances of 0.0005 -

0.002". Let's use the large value which is the w in the above formula. A 4" bore isn't all that uncommon so we'll use that which gives an r of 2". I didn't find any listings for distance from the piston groove to the top of the piston, but pistons I've seen in the past were probably around 0.25" so I'll use that for h. This gives a volume of 6.28 x 10^-4 cubic inches which is miniscule in my book.

Makes little difference. Let's assume the above cylinder has a stroke of 3.5". The volume swept by the piston is therefore: pi*r^2*h or 44 cubic inches if I punched the calculators buttons correctly. If the compression ratio is about 10:1, that gives a cylinder volume of approximately 5 cubic inches. This means that the volume of cylindrical shell above the top ring amounts to ~0.001%. So even doubling or halving the diameter of the engine bore will make diddly squat difference and wouldn't contribute enough to exhaust emissions to be even measureable.

Have you been embarrassed enough yet? Can you find any flaws in my assumptions or calculations or, better yet, provide some calculations of your own that support your theory about this contribution to engine emissions?

And the above percentage is assuming that the entire volume of gases trapped in this annulus is exhausted unburned on each cycle. Anyone with a modicum of knowledge of fluid mechanics knows that you'll have a very hard time moving the air/fuel mixture into and out of this cylindrical shell that is only 0.002" thick or less. So the effective percentage above is probably more like 0.00001%.

Matt

Reply to
Matthew S. Whiting
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Joe, cat got your tongue? :-)

Matt

Reply to
Matt Whiting

A little bit of information for anyone following this thread.

Stochiometric ratio is 14.7 to 1. A little leaner is better for emissions. A little richer better for power. If you could get it to fire at 25 to 1, the cylinder would get hot enough to melt your pistons.

Peak torque is always lower than peak horsepower. Generally the larger the engine the lower the RPM at which the torque peak is reached. Generally on an unmodified engine, big displacement V8 will be in the 2400 RPM range, on a small 4 cylinder it will be around 3600 RPM.

At peak torque, the engine also has the highest brake specific fuel consumption. In other words, the engine is operating at it's most efficient level. Most power for the amount of fuel, that's it. If it is operated at a lower RPM it will get better fuel mileage simply because it is not drawing as much fuel in for the distance traveled.

A smaller cylinder is easier to control emissions because it is easier to maintain density of mixture in a smaller area, thus preventing hot and cold spots. You have to take into account the amount of time it takes the flame to travel across the cylinder. On a four-stroke engine at 2000 RPM, if you had the full 90 degrees of the power stroke to burn the fuel, that would be .06 seconds to burn the entire mixture (60 seconds/(2000 RPM/2 Revolutions for a full cycle of the engine)). At 4000 RPM that becomes .03 seconds. The speed of the flame remains constant. So, if you increase the volume dramatically, you will never burn the entire charge. This is why the ignition timing changes with engine speed.

Reply to
A. Cheshire

In this newsgroup everybody needs at least 350 HP to win the Traffic Lights Grand Prix... :-)

DAS

Reply to
Dori A Schmetterling

Good info. Thanks!

Bill Putney (to reply by e-mail, replace the last letter of the alphabet in my address with "x")

Reply to
Bill Putney

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