Okay, so I was reading on how a four stroke engine works and was watching a video on how a car transmission system works. Here are the links:http://ift.tt/1vb5BVH
http://ift.tt/1vb5BVI
Now, after having read about the engine and watched the video, there are a handful of things that aren't very clear to me.
Firstly, regarding the engine, precisely what starts the intake cycle(i.e what triggers the piston to descend from the top to the bottom of the cylinder)? From my understanding, if it is the fuel, then surely the fuel would need to undergo combustion first in order to produce thermal energy which would then be harnessed to do mechanical work on the piston. But since step 3(combustion) has not been reached yet at this point, how does the fuel trigger the motion of the piston, if it is indeed the fuel that is triggering it? The other thing that confused me was the multicoloured P-V diagram next to the 'Fuel Considerations' part of the first link. It says 'the intake(A) stroke is performed by an isobaric expansion'. Now what I understand is that in the intake cycle, the downward motion of the piston increases the volume of the cylinder(and consequently the volume of the gas) and assuming that it is an ideal gas at a constant temperature, by Boyle's law, its pressure decreases. So why, then, is it called an isobaric expansion?
My other doubts concern the video in the second link, specifically the gearbox. Now the gears on the main shaft are all in motion and the dog clutch will need to engage with the intended gear to induce the main shaft to rotate. Firstly, what induces the gears to rotate? Secondly, when the diameter of the gear to which the dog clutch is engaged decreases, the speed of the engine increases while the torque decreases. My understanding is as follows: all the gears are rotating at the same angular speed. Torque, τ, is the product of the force, F, and the perpendicular distance between the force and the centre of the axis on which the force acts, r. When F is constant, τ is directly proportional to r. But as it relates to the inverse proportionality relationship between the speed of the engine and the diameter of the gear, I'm quite in the dark. Does it have to do with the notion that the rate of production of thermal energy by the engine(which is harnessed as mechanical power, P) is constant(as the pressure applied by the driver's foot on the accelerator is constant), implying that the angular speed, ω, of the main shaft, is inversely proportional to the torque since P= τω?
I would appreciate it if anyone could please tell me if my understanding is correct and explain the bits which are not clear to me or the bits which I've understood incorrectly. Please also feel free to add anything interesting to the discussion.
http://ift.tt/1vb5BVI
Now, after having read about the engine and watched the video, there are a handful of things that aren't very clear to me.
Firstly, regarding the engine, precisely what starts the intake cycle(i.e what triggers the piston to descend from the top to the bottom of the cylinder)? From my understanding, if it is the fuel, then surely the fuel would need to undergo combustion first in order to produce thermal energy which would then be harnessed to do mechanical work on the piston. But since step 3(combustion) has not been reached yet at this point, how does the fuel trigger the motion of the piston, if it is indeed the fuel that is triggering it? The other thing that confused me was the multicoloured P-V diagram next to the 'Fuel Considerations' part of the first link. It says 'the intake(A) stroke is performed by an isobaric expansion'. Now what I understand is that in the intake cycle, the downward motion of the piston increases the volume of the cylinder(and consequently the volume of the gas) and assuming that it is an ideal gas at a constant temperature, by Boyle's law, its pressure decreases. So why, then, is it called an isobaric expansion?
My other doubts concern the video in the second link, specifically the gearbox. Now the gears on the main shaft are all in motion and the dog clutch will need to engage with the intended gear to induce the main shaft to rotate. Firstly, what induces the gears to rotate? Secondly, when the diameter of the gear to which the dog clutch is engaged decreases, the speed of the engine increases while the torque decreases. My understanding is as follows: all the gears are rotating at the same angular speed. Torque, τ, is the product of the force, F, and the perpendicular distance between the force and the centre of the axis on which the force acts, r. When F is constant, τ is directly proportional to r. But as it relates to the inverse proportionality relationship between the speed of the engine and the diameter of the gear, I'm quite in the dark. Does it have to do with the notion that the rate of production of thermal energy by the engine(which is harnessed as mechanical power, P) is constant(as the pressure applied by the driver's foot on the accelerator is constant), implying that the angular speed, ω, of the main shaft, is inversely proportional to the torque since P= τω?
I would appreciate it if anyone could please tell me if my understanding is correct and explain the bits which are not clear to me or the bits which I've understood incorrectly. Please also feel free to add anything interesting to the discussion.
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