great read as always Jmac
great read as always Jmac
MdOiDcEtRaAtToOrR
1996 Chevrolet Impala SS 14.5 @ 94
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just subscribing so i can read it later
welcome back James.
Well done.
What the f00k happened to this place?
Another good read.
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Originally Posted by DJGTSR
While I can't/dont want to read it all right now, still is good ****.
Good info, I never knew the forumula for displacement
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couldn't have said it better myself! you ASE cert? Should be if your not.
You need torque to take off and then you need hp to keep you moving.
One of K&N commercials claims it takes somewhere around 17 hp to propel a v8 truck down the highway @ 55 mph
Low to the earth and stanced
In a 4 cylinder engine, 1 cylinder will be in 1 of the given cycles at any given time. This means that 1 cylinder will provide torque every ¼ revolution. In an 8 cylinder engine, 2 cylinders will be in 1 of the given cycles at any given time. This means that 2 cylinders will provide torque every ¼ revolution which means that the torque from each of those two cylinders is combined to turn the crankshaft. So, in our 4 cylinder engine, we’re making 209.44 lb-ft of Torque while we’re making 418.88 lb-ft of Torque in the V8.
I dont get this.. if the 4 cyl makes 209tq every power stroke, then it would make 418 in a revolution, because it has 2 power strokes every revolution.
You are saying that a V8 would fire all 8 cylinders in a revolution, but this isn't true.
1 cyl in a 4cyl doesnt fire every 1/4 rev, I dont understand this.
360/number of cylinders *2 =how many degrees it fires in crankshaft rotation
360/4 *2=180. A 4cyl will fire every 180degrees, not every 90. A V8 wont fire 2 cyls at the same time, so 2 wouldnt be every 1/4 rev.
a 400hp 4cyl will fire 2 times in a revolution, and a 8cyl 400hp car will fire 4 times in a rev, both will produce 200hp, because in the 4cyl each cyl produces 100hp, and in the 8cyl each produces 50hp. So if each cyl in a car makes 209hp,and the car is a 4cyl, it would make 418hp per revolution,not 209.
Okay, so the engine produces Torque, but how do we transmit that Torque to the ground ? Well, on the way to the ground, it has to pass through the transmission, differential, and wheels to get to the ground. Let’s say we have a first gear ratio of 3:1, a final drive (differential) ratio of 4:1, and a set of 195/60/15 drive wheels. Okay, so our engine is churning out 209.44 lb-ft at the crank in our 4 cylinder engine and the first gear has a Radius 3 times greater than that of the crankshaft’s, so we get 209.44/X = Y/3X … If we multiply each side by 3X, we get 209.44*3 = 628.32 lb-ft of Torque … So the gear pretty much acted as a Torque multiplier. Now we pass down the driveshaft to the differential where we have a ratio of 4:1 or 4 times greater Radius than First gear, so we get 628.32/X = Y/4X … 628.32*4 = 2513.28 lb-ft of Torque … Again, it acted as a multiplier. Now we turn to the wheels, where we actually have a known Radius and a known Torque, but we don’t know the Force. To get the diameter of the wheel, we take the rim size (15”) and add the height of the tire (2*195*0.6/25.4) = 24.2126” or a Radius of 12.1063” … 1 foot = 12” and 2513.28 lb-ft is equal to applying 2513.28 lbs of Force to a 1 foot Radius or X Force applied to a 12.1063/12 Radius … So we get 2513.28*1 = X*12.1063/12 … So we multiply each side by 12/12.1063 and get X = 2513.28*12/12.1063 = 2491.212 lbs of Force … So we finally translated our engine’s torque into a force at the wheels. So this means that Torque is all that matters, right ? Wrong …
Can you explain this a little more for us stupid people
jesus. 0_0
ill read it later
math is hard.
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