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Why Does My Differential Not Give Power To Both Sides

In a rear wheel drive car, why is a differential provided at the rear if turning is done by front wheels?

A nice question.The confusion arises form the fact that many a times the textbooks and other sources say this.Q. Why is a differential required?A. A differential is required because of turning.The correct fact and the wording should be like this. In any vehicle when the drive is given to two wheels one on the LEFT and another on the RIGHT we need to have a mechanism to ALLOW  these two wheels to turn at different speeds* and still transfer power to them. That mechanism is differential.Steering is the mechanism by which we actually turn the vehicle.The steered wheels could be same as/different than driven wheels however the differential is always needed on the driven wheels. The word "ALLOW" is the keyword here for clear  understanding.*The need for different speeds is well documented and and illustrated.Edit : OK Now if the differential only ALLOWS them to rotate at different speed the the question arises as to what MAKES them rotate at different speeds. The answer is nothing within the car mechanisms makes them  do that. It is the road contact and the directional movement of the vehicle as dictated by the steering wheels that MAKES them do it.

How do I solve this differential equation (1+y) dx - (1+x) dy = 0?

This is a textbook example of a separable differential equation.The equation given is: [math](1 + y)dx - (1 + x)dy = 0[/math]This can also be written as: [math](1 + y)dx = (1 + x)dy[/math]Therefore: [math]\frac{dx}{1 + x} = \frac{dy}{1 + y}[/math]We can integrate both sides: [math]{\displaystyle \int} \frac{dx}{1 + x} = {\displaystyle \int} \frac{dy}{1 + y}[/math][math]ln(1 + x) + C = ln(1 + y)[/math]Take [math]e[/math] to the power of both sides: [math]e^{ln(1 + x) + c_1} = e^{ln(1 + y)}[/math][math]e^{c_1}(1 + x) = 1 + y[/math][math]y = c_2x + (c_2 - 1), c_2 = e^{c_1}[/math]Since we have no need for further manipulations, I will call the constant [math]c_2[/math] as [math]K[/math], so: [math]y = Kx + (K - 1)[/math], [math]K[/math] being a real constant.

Transmission (mechanics): In an open differential, an engine sends the same amount of torque to both the wheels. The same amount of torque would mean same speed. How does this differ from the locking differential?

Engine sends energy, in form of rotational force i.e. torque, to the differential through transmission. Differential distributes the torque.[Source]In open differential the distribution is fixed at 50-50. Both the wheels get the same amount of torque. Even when the two wheels are rotating at different speed, their average speed is always the same, equal to the speed of crown gear.In locked differential however, the differential gears are locked together in certain cases i.e. when traction available to both the wheels differs drastically. This makes differential behave as a solid axle, forcing both the wheels to rotate at the same speed. As traction available to the wheels is different and yet the wheels are rotating at same speeds, the torque distribution is no more equal to 50-50.This feature makes the locking differentials desirable in rough terrains, where half the time one of the wheels is in air or mud or water. When one of the wheels is in air, open differential will provide zero torque to both the wheels whereas locked differential will provide more torque to the wheel which is on the ground.

If we have an airplane with two engines, and we divert the power to one of the engines to turn the aircraft, how does it compare with thrust vectoring, and what do we call it?

First, you cannot “divert” the power of one engine to the other one, jet or piston. This isn’t Star Trek after all. You can put the aircraft into cross feed, that would feed the right engine with the left engines fuel (and vice versa) but that wouldn’t divert any power, it would only give you a longer run time for that engine. The only thing you can do on a twin would be to increase or reduce the percentage of power on one engine, which would be called asymmetrical thrust (a Cessna 337 push-me-pull-you not withstanding). I had an engine failure on a PA-31 just before I was cleared for take off (someone was certainly looking out for me that day) and had to taxi back to the hanger on the remaining engine. The asymmetric thrust was so bad that I had a very difficult time getting back. Too much power and the engine would pull the plane off course no matter how much nose wheel steering and asymmetric braking I put in; not enough and it would slow to a stop. It was very hard to get it rolling and then maintain just the right amount of power to keep it moving under control. The most amazing story about using asymmetric power is the United flight 232 crash in Sioux City. Absolutely incredible, look it up.

How do you put limited slip differential on a car? Why?

When you are driving round a corner, the inside wheel has to turn slower than the outer wheel because it is following a tighter radius than the outer. Without a differential, the wheels will fight each other and will skid. This makes controlling the vehicle more difficult and wears out the tires.But a differential has an inherent design problem. When you try to start moving on a slippery road and one wheel starts to slip, the differential transfers more power to the wheel that is spinning and less to the one that has grip. The faster the slipping wheel turns, the less power gets transferred to the wheel that grips. This means you wind up going nowhere until something happens to start equalizing the grip on both wheels. For example, throwing sand under the slipping wheel to increase traction.A limited slip differential has a clutch that links the inside and outside wheels and transfers some of the power between the slipping and gripping wheel. So, on a slippery road where one wheel has less grip, you have a chance of moving with a limited slip differential.As far as putting a limited strip differential on a car, there are two situations. If the car has a live axle, such as an old rear-wheel drive american car, you will need to remove the entire rear axle and differential assembly and replace it with one that has a limited slip clutch. On some old american cars, a limited-slip differential was an option you could buy. So you might be able to find one in a wrecking yard. If the car has front-wheel drive, the differential is integrated with the transmission and not something readily available. So, I am afraid it won’t be possible.

How is engine power delivered to the wheels of a front wheel drive car?

Thanks for the A2A.Drive is transferred from the differential mounted within the gearbox through two shafts. These shafts each have two flexible joints on them, to allow for suspension movement and steering.They’re called constant velocity or CV joints.A cut-away:These devices allow the output side of the joint to turn at the same rate as the input side. If Universal Joints were used, the vehicle would suffer from excessive vibration, as the output from a universal joint is not linear, they accelerate & decelerate reletive to the input.Universal joint also only operate over a narrow range of deflections, which would allow for a very poor steering lock if they were used to transmit drive in a front-wheel drive car.(Above). Universal joints fitted to a rear-wheel drive system.Universal joints are not a problem when used to transmit drive from a front mounted engine to a rear axle, as the angle through which a pair of joints move can will be identical, plus the joints can be staggered out-of-phase so as to cancel-out any vibration due to the acceleration/deceleration of the yoke.So, universal joints are not used in front wheel drive vehicles. CV joints are used instead.A schematic of a front wheel drive system:The driveshafts are shown in red. The inner ends are attached to the output flanges on the gearbox (not shown). The cv joints are housed within the flexible rubber boots on the driveshafts.

Differential Equation help...nonhomogenous undetermined coefficients?

y''+3y=-48x² e(^3x)
So you guessed there is a solution of the form :
y = ax² + bx + c which gives:
y’ =2ax + b
y” = 2a
Checking whether this is a solution we write:
3ax² +3bx +3c +2a = -48x².e^(3x)

Now the coefficients of each power of x and the constant must be the same on both sides of this last equation. This is not possible due to the exponential term which you seem completely to have dropped in your expansion. So with your trial solution you’re barking up the wrong tree! I’ll take a look at getting a better solution for you and edit my answer if I succeed.

{If I were to take your expansion as written it gives a = -48/3 = -16 ; b=0 and c= -⅔.a = 32/3
So y = -16x² +32/3
y’ = -32x
y” = -32
This would obviously provide a solution, were it not for the exponential function on the right side!}

EDIT:
Try: y = [ax² + bx + c].E where E = e^(3x)
y’ = 3.[ax² + bx + c].E + [2ax+b].E
y” = 9.[ax² + bx + c].E + 3[2ax+b].E + 3[2ax+b].E + [2a].E
= E.[x²(9a) +x(9b+12a) + (9c+6b+2a)]
So the equation gives:
y’ + 3y = E.[x²(12a) +x(12b+12a) + (12c+6b+2a)]

Giving, by equating coefficients of x²: 12a = -48 or a =-4
by equating coefficients of x: b =-a = 4
and by equating the constants: 12c = -(24 – 8)/12 = -4/3

Why does the left wheel of a car in air rotate in reverse direction when the right wheel is rotated in one direction?

Research the operation of something called differential gearing. The driven wheels of a vehicle must be able to rotate at slightly different speeds when the vehicle goes round corners, otherwise the tyres would scrub off their tread abnormally fast. Differential gearing allows this.One of the side effects of this gearing is when the vehicle is jacked up and the engine is off and out of gear turning one driven wheel causes the opposite wheel to rotate in the reverse direction.

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