Displacement, Compression ratio, horsepower, torque, thrust and shift points. WARNING: THIS IS DAMN LONG, BUT GOOD (I may be biased) Can some of you weld like none other but have trouble multiplying? Can you rebuild an engine but have trouble figuring out the new compression ratio? Well if any of these sounds like you read up and bask in your newfound smartness. 3.14 will be substituted for pi, if you want a more precise number you can use 3.1415927. ^2 means some number squared. And / represents division, 6/2 would be 3. Make sense? Good. Ready? Displacement Displacement = (3.14 x bore^2 x stroke x # of cylinders) / 4 For example, a Chevy 350: 3.14 x 4^2 x 3.48 x 8 = 1398.6816 1398.6816 / 4 = 349.6704, which rounds to 350! Compression Ratio Compression Ratio = (Cylinder volume + chamber volume) / chamber volume This is a lot trickier to figure out, you will need several numbers for this. First is the cylinder volume, which is very similar to our displacement formula: (All standard measurements must be converted to metric or vise-versa. Use the conversion factor 2.54 to convert your bore and stroke measurement to cm BEFORE you start multiplying things together.) (3.14 x bore^2 x stroke) / 4 Which equates to a 716.259cc single cylinder volume in our 350. The next number you will need is the chamber volume; this is the volume of the compression chamber when the piston is at top dead center. One way to do this is with the engine****embled, tilt it so the spark plugs on one side are vertical, pour liquid from a measured container into the cylinder until the top of the liquid reaches the start of the spark plug threads, subtract the amount of liquid remaining in the container from the amount you started out with and that is your chamber volume. For example, if you have a container with 100cc of gasoline and after pouring it into the cylinder you end up with 20cc in the container 100 20 = 80, so your chamber volume is 80cc. Another way is to add up known volumes, say you buy some Edelbrock Performer RPM heads for your 350 with a chamber volume of 64cc and some Speed-Pro POWERFORGED pistons with a dome of .200in. or .508cc. Now you must find the thickness of your head gasket. Say your gasket has a thickness of .045in. or .1143cc (if at top dead center your pistons do not come all the way to the deck now would be the perfect time to add that distance) using the formula (3.14 x bore^2 x gasket thickness) / 4 we can conclude that our gasket adds about 9.262cc. Ok, lets get all our numbers together. 64cc head chamber volume .508cc piston dome volume 9.262cc gasket area volume We get our chamber volume using Chamber volume = head volume + gasket volume piston dome volume (If the piston dome were concave, then you would add the piston dome volume) This equates to a 72.754cc Chamber volume. Now we can move on to the rest of our equation. (Cylinder volume + chamber volume) / chamber volume (716.259 + 72.754) / 72.754 This gives us a compression ration of about 10.85, race gas anyone? Horsepower What is horsepower exactly? It is the ability to apply leverage (torque) over a certain amount of time. Basically torque over time. An engine producing a certain amount of torque at a high RPM will have more horsepower than an engine producing the same amount of torque at a lower RPM, because at the higher RPM its doing more work. 1 horsepower is equal to 550 pound-feet per second. If you want all the gritty details about how that number came about or how the next equations were made get the Auto Math Handbook theyll tell you. Horsepower = (rpm x torque) / 5252 (5252 is constant) Lets say our 350 makes 500 lb/ft of torque (its got a 10.85 compression ratio for Christs sake!!!) at 3000 rpm. (3000 x 500) / 5252 Thats only 285.6 hp @ 3000 rpm. Why dont we move that 500 lb/ft up to 6000 rpm (6000 x 500) / 5252 Now weve got 581.2 hp @ 6000 rpm, not too shabby!! Torque Im not gonna go in-depth in to torque, cause weve all used a torque wrench before, so heres the equation: Torque = (5252 x horsepower) / rpm Wheel Thrust & Wheel Torque You cant measure your cars output in thrust!! Thats for airplanes and the space shuttle!! Actually you can, and heres how. Lets****ume your car is putting out 300 ft/lbs of torque and you have a Warner T-10 4 spd. With a 2.20 1st gear ratio, and in your 9-inch rear you just got a set of 4.10s, Yeah! Basically to get your torque at the wheels you multiply your engine torque by any gear reductions you may have, like the******* and rearend. So in this case in 1st gear: 300 x 2.20 x 4.10 = 2706 lb/ft at the wheels. What does this have to do with thrust? Read on HAMB brethren were getting to that! Essentially the tires on our car are our lever between the axle and the ground, if you have 24 in. diameter tires that is a distance of 12 in. from the center of the axle to the ground. If we our making 2706 lb/ft at the wheels, with a 1 foot long lever to the ground, then that is 2706 lbs of force to the pavement, or 2706 lbs of thrust, YEAH!!! But say we put some big 31 in. cheaters out back, now what do we do? Simple: Wheel Thrust = Wheel Torque x 12 / Wheel Radius Sooooo 2706 x 12 / 15.5 = 2096 lbs of thrust. Which explains why putting larger tires on will decrease your acceleration, because you have less thrust. Ideal Shift Points This will be somewhat short and sweet cause Im tired, when you are racing and you shift, you want the rpm that the engine is it right after the shift to be the rpm where your engine is making maximum TORQUE, not horsepower. Torque is what accelerates you. Ideal shift rpm = (ratio shift from / ration shift into) x max torque rpm So the first gear on our Warner T-10 is 2.20 and the second gear 1.66 and lets say our engine is making max torque at 4000 rpm (2.20 / 1.66) x 4000 = 5301 rpm, so that is the rpm for your 1st to 2nd shift. Hope that wasnt to boring, and you can find most of that stuff in the Auto Math Handbook by John Lawlor, read it, live it. -Eric
Flexi, This is awesom! 5 stars for you! I'm a newbie so I like things like this I can understand and apply! Danny
Good stuff! I had a math teacher back in high school that took the time to help me understand algebra by relating it to cars, since he knew I was a gearhead! We went through all those formulas and more! He was really incredible, we'd be sitting there, and I'd say something like "Suppose I want to figure out what rear gears I'd need to get my car through all three gears in a quarter mile and be right at the 6200rpm redline at the finish line...could you figure THAT out?" He didn't even blink...just told me to bring him a measurement of my rear tire diameter, gear ratios in the trans and rear axle...and we'd figure it out the next day! I don't remember all of the specifics, but I was impressed that he was able to use math to help me build my car when he didn't even change his own oil! He was also the one wo taught me to think of a vehicle as an equation...with the engine and drivetrain as smaller equations inside of the big equation! He said that cars come off the****embly line "balanced"...and if you want to modify one successfully, you have to understand that every ONE change you make affects the whole equation. If you make a change over HERE, then you must make an equal change over THERE to keep it balanced! You can't add a bigger cam without being able to flow more air and fuel...you can't go to taller gears without the ability to increase your engine's maximum rpm...or the rpm at which peak horsepower and torque are reached...and so on! Funny, that a little Russian math teacher who knew next to nothing about cars was able to teach me MORE about building a car than my shop teacher did! It's more than nuts and bolts...there's math and science involved!
Glad you guys like it , I think cars are the only reason I like math so much. The first thing I did when I learned trig last year was to go home and figure what the new driveshaft angle for my jeep would be and how much suspension travel I would have after i lifted it. You can do anything with numbers! Also, the "Auto Math Handbook" by John Lawlor is where I got alot of those equations, he is a genius, In the book there is also center of gravity, weight transfer, quarter mile times, and blood alchohol level. It is a good book!
Well damn man, thanks for all of that...I'm still tryin to figure a lot of it out in my head, but its good stuff!
Every magazine editor in the country has that book, and we use the hell out of it! It should be mandatory in everyone's library. Excellent, excellent book. -Brad
I used his formula's to calculate the right tire size to run in the Quarter mile. I knew my power curves from my dyno sheet and the gearset in my Quick change, so I did a little more fine tuning with the Tire size! Grumpy Jenkins and Smokey Unik always used formulas and they were my first mentors in racing! Cool Stuff!!! Mark
[ QUOTE ] Displacement = (3.14 x ... [/ QUOTE ] " 3.14 will be substituted for pi, ^2 means some number squared. And / represents division " Not trying to nit pick,but: The DISPLACEMENT of a motors is the VOLUME of each cylinder,multiplied by the number of cylinders.Right? The CIRCUMFERENCE of a circle is Pi x D,where D is the diameter. The RADIUS of a circle is 1/2 the diameter. The AREA of a circle is Pi x r^2 ,where r is the radius,or in this case (Bore/2). The VOLUME of any cylinder is Area x Height.Which would be Pi x (bore/2)^2 x stroke So the displacement of a motor = Pi x r^2 x stroke x # cyl = 3.14 x (bore/2)^2 x stroke x number of cylinders = 3.14 x 2^2 x 3.48 x 8 = 12.56 x 3.48 x 8 = 349.6704 You get the same answer with both formulas,in this example, only because the bore^2 is canceled out by the /4 at the end. bore^2/4= 4^2/4 = 16/4 = 4 while it should be (bore/2)^2 (4/2)^2 = 4 You would get exactly the same answer if you used the formula: Pi x bore x stroke x #cylinders,for the same reason. Why does this make a difference ? Because,if the bore is any other number,than 4,you will get the wrong answer. Circle formula Area and volume formulas Volume formulas Automotive calculators.
While that makes sense, I don't think its true. Correct me if I'm wrong for any number x: x^2/4 = (x/2)^2 8^2/4 = 64/4 = 16 (8/2)^2 = 4^2 = 16 3.75^2/4 = 14.0625/4 = 3.515625 (3.75/2)^2 = 1.785^2 = 3.515625 You get the same number either way. I do however appreciate the criticism, I could very well have been wrong.
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