great read as always Jmac
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great read as always Jmac
just subscribing so i can read it later
welcome back James.
Well done.
Another good read.
While I can't/dont want to read it all right now, still is good ****.
Good info, I never knew the forumula for displacement
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
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 :p:
math is hard.