need a little help with some springs

[Wood_workur]

I'm planning a jewelry box much like this:



But what I want to do on the inside is have a heart on the lid like this: http://www.bt3central.com/showthread.php?t=27245 but the thing about that is the left (or right) side of the heart is gonna be the release to a secret compartment- when you squeeze that side up to the other side, its gonna pull two dowel rods out of the side that will let you pull the fake top out and open the compartment.

The thing is, I'm not sure how strong to make the springs that hold the dowel rods out normally. I was thinking that a total of 5 lbs to release the entire thing would be sufficient without making it impossible to open (I don't want the release to be obvious, but I don't want it accessible to superman only), and that with 1/4" of compression and about 1/2" of spring compression in the open position that I'd need a spring with at least 3/4" compression length, and since I'm using two I'd need 2.5 lbs each at max compression which is 3/4" to have a spring that compresses at about 3.33 lbs/inch, am I correct? Do the numbers I'm using make sense?

And of course, where can I buy these springs? The two places I found want to change me extra for not ordering in bulk and want like $4 a spring.

[LCHIEN]

Hardware storres like Ace Hardware usually have a section nears the small parts with a large selection of springs. You'll find all diameters and lengths and spring constants.

if you recall from hig school physics, springs have a spring constant, K in lbs per inch or ounces per inch. The force the pring applies is proportional to the compression distance D, F=K x D. Up until the spring bottoms out, of course.

[chopnhack]

LChien, doesn't the force become greater as the spring gets compressed? Or is it negligible in smaller/shorter springs?

[Daryl]

You can also get a variety pack of springs for just a few bucks. If you don't want to do Loring's physics calculations you can just start subbing springs from the pack until you come across what you like.
Beautiful box, that wood looks like it is in 3-D.

[Wood_workur]

LChien, doesn't the force become greater as the spring gets compressed? Or is it negligible in smaller/shorter springs?

Work increases exponentially. Force is linear. F=kx, and if you integrate that you get W=.5kx^2

Basically what I want to know if the idea of 5 lbs of force to open the hidden compartment is a reasonable amount of pressure. I've got no real way of measuring it, and googing didn't turn anything up.

[chopnhack]

Thats one of those "feel's right" questions that I think you may only resolve by building a quick and dirty mock up. That way you can try some options and see what works for you. It will also be good practice for the real piece.

[LCHIEN]

from it's "rest" point, if you compress is it takes force to compress it and once compressed it will push back with that same amount of force when held in place.
Likewise if you pull it with force, it will elongate and want to pull back withthe same amount of force.

Chop, 5 lbs is a lot of force. Consider a gallon of milk weighs 7.3 pounds roughly. so 2/3 full gallon will be about 5 lbs. If you tie a string to the drawer and hang the milk bottle off the side it will rip the drawer out so I thjink that's a bit on the high side. Assuming relatively low friction I would guess a starting point would be about 6 oz of force... less than half a pound and that might be high. The thing to remember is that as the spring relaxes the force will go down. Unlike the milk bottle which will exert 5 pounds all the way until it hits the floor, a spring will start with 6 oz but at the end of its travel it will have zero force. The amount of work it will do is the force over the distance. Do you just want to pop it open an inch to get a hold of it with a finger or do you want a cash register in which it pops all the way open?

[Wood_workur]

Do you just want to pop it open an inch to get a hold of it with a finger or do you want a cash register in which it pops all the way open?

You see the heart on the inside of the top of that jewelry box? What I'm doing is taking one of those sides, not having it glued, but rather putting a piece that goes through the fake lid. Into that goes two dowels that go through the top from one side to where those pieces are. Just past those pieces is where I'm gonna put the spring, and its job its to provide a bit of resistance to moving one side of the heart, and to make it spring back once it has been pressed. That is why I want to always have it under some compression, so that it will always return to the extended (as in the dowels that lock it in place are extended out) position.

So in that situation, what would you put in there? I was messing around with a bathroom scale, and I managed to get it to 15 lbs consistently. I figure I wouldn't want it under that much tension, and I'm not sure how much is enough without being too much.

[LCHIEN]

if i read you correctly the heart protrusion will slide sideways. You just need enough tension to make it slide back into normal position so I'm thinking a few ozs. of force, Yes you should have some preload on the spring, otherwise the force at normal position will be zero and not really snap back as it should.

[Wood_workur]

if i read you correctly the heart protrusion will slide sideways. You just need enough tension to make it slide back into normal position so I'm thinking a few ozs. of force, Yes you should have some preload on the spring, otherwise the force at normal position will be zero and not really snap back as it should.

okay. I'll look at getting just under a pound in there.