engineering question.

[Adam32]

I had a clutch fork in a transfer box which was plastic. It broke so I had a custom metal one made. The plastic one weighed 100grams and the replacement metal one weighs 600 grams.

Now there is a spring which is mounted to the inside of the case and pushes against the fork, so that when you change back down to 2 wheel drive, the fork slides down and disengages a clutch plate. Without the spring, the fork could bind in place and you wold be stuck in 4 wheel drive. This is especially true, as the fork is not aided by gravity - sitting on it's side when the transfer box is mounted to the vehicle.

So the question is what compression strength of spring should I go for as a multiple of the existing spring? As the metal fork is heavier than the original plastic, the spring will obviously need to overcome a greater force. But do I really need something with 6 times as much compression strength? During bench testing the original spring designed for the plastic seems to work. But I feel to be on the safe side, I should fit a stronger spring. I don't want to put the box back on the vehicle, only to find out the original spring is not up to it and then have all the time / messing about to pull the box off again (especially as I work outside in the cold!).

So maybe I should try something with twice as much compression strength as the original spring and don't really need 6 times as strong, considering that even the original seems to work in bench testing.

Any thoughts would be appreciated.

[CrutyRammers]

If it works on it's side (I presume it pivots in the middle somewhere), I don't see that the extra mass will come into it at all really.

[TooMany2cvs]

Is the spring mostly overcoming the weight of the fork - or other forces inside the box?

Or is it simply holding the fork out of the way?

[tapkaJohnD]

The fundamental frequency fo a mass-spring system is given by

F = (1/2Pi) . SQR(K/M) K = spring stiffness M = Mass

So if the mass is six time greater, the stiffness should alo be six times greater to achieve the same frequency.
BUT, the spring-mass is not oscillating freely, but being compressed and then released.

The velocity of an object propelled by a compressed spring is given by

V = X . SQR(K/M) where X is the compressed length of the spring

So again, to achieve the same velocity, the spring stiffness must increase with the mass.

But Velocity is distance/time, and the less stiff spring will move the fork more slowly If you are content to wait a moment longer when changing between four and two wheel drive, then you will probably be OK.

JOhn

[Penelope Stopit]

tapkaJohnD said:

The fundamental frequency fo a mass-spring system is given by

F = (1/2Pi) . SQR(K/M) K = spring stiffness M = Mass

So if the mass is six time greater, the stiffness should alo be six times greater to achieve the same frequency.
BUT, the spring-mass is not oscillating freely, but being compressed and then released.

The velocity of an object propelled by a compressed spring is given by

V = X . SQR(K/M) where X is the compressed length of the spring

So again, to achieve the same velocity, the spring stiffness must increase with the mass.

But Velocity is distance/time, and the less stiff spring will move the fork more slowly If you are content to wait a moment longer when changing between four and two wheel drive, then you will probably be OK.

JOhn

Brilliant post, I enjoyed every moment of my read.

Edited by Penelope Stopit on Monday 12th February 10:17

[brrapp]

Mass of the lever isn't the same thing as the force the spring is pushing against. The lever is probably close to being in balance so the mass of the lever probably won't make a difference. I can't see anyway of working out from a pure maths solution. Either try to measure the force required to pull back the lever when its in place (maybe with a spring balance or similar), or just try it with the original spring and see if it works.

[CrutyRammers]

brrapp said:

Mass of the lever isn't the same thing as the force the spring is pushing against. The lever is probably close to being in balance so the mass of the lever probably won't make a difference. I can't see anyway of working out from a pure maths solution. Either try to measure the force required to pull back the lever when its in place (maybe with a spring balance or similar), or just try it with the original spring and see if it works.

^^^ That.

[Adam32]

tapkaJohnD said:

So again, to achieve the same velocity, the spring stiffness must increase with the mass.

But Velocity is distance/time, and the less stiff spring will move the fork more slowly If you are content to wait a moment longer when changing between four and two wheel drive, then you will probably be OK.

JOhn

Thank you so much for your explanation. So as long as my current spring is managing to overcome the weight of the object, then the only thing that will change is the speed at which the fork moves.

I guess the only other consideration is when the box is on the vehicle it is full of oil. So obviously there will be the added pressure of the oil to push against. But I guess that is not much?