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31st December 2014, 12:42 PM #1
I need a person with a sharp brain
I have a new tool for my CNC.
It is a springloaded engraver.
Image 6.jpg
I need to work out how much to turn the screw (1) to equal a force of 500 gram upwards of the plate (10)
I do not need to know how much weight to pull the plate down, I need the upwards pressure (resistance) on the spring.
In other words, that is the pressure I need for the bit to follow any unevenness in the material.
The plate will be resting on he top of the bottom (u-shaped) part of the tool.
The router is resting in the hole (10)Every day is better than yesterday
Cheers
SAISAY
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31st December 2014, 01:20 PM #2Retired
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Why not put a small gram-accurate scale underneath the left hand component of the probe and adjust until it reads 500?
I'm assuming the right hand is dynamic and the left static, so as the assembly descends it lifts the right upwards as its sprung downwards. If the assembly has a motor in it, it will weigh more?
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31st December 2014, 04:32 PM #3Banned
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Isn't it just a matter of trial and error. If you place a 0.5kg weight on the right hand side it will hit the bottom of the spring is not screwed in enough or not the case may be.
Apply weight, observe, remove weight and adjust screw abs repeat.
It depends on how accurate you need it but I would have thought you would get close enough.
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31st December 2014, 06:40 PM #4
I think you misunderstand me.
Measurements and Instructions.jpg2444_0.jpg
I need to know how much to tighten the screw to lift the plate with a 500 gram upwards pressureEvery day is better than yesterday
Cheers
SAISAY
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31st December 2014, 07:08 PM #5
I may be misreading this but here's my solution.
You will need a standard weight, something appropriate to the spring, or you will need a scale/balance. You will also need a ruler.
Take the spring off of the mount and place it on a hard surface and put the weight on top. Measure the height of the spring. Now remove the weight. Measure the unloaded height of the spring. For springs within their linear (elastic) range F = -kx where F is the mass times the acceleration of gravity, constant in this case, and x is the change in length of the spring.
Here's a hypothetical. Say you put a 100 gram weight on top of your spring and measure that it's compressed length is 5 cm. When you remove the weight the spring length is 6 cm. Your spring constant, k, is 100 gm/(6cm -5cm) or 100gm/cm. Thus if you want the spring to exert 500 gm of force you need to compress it 5 cm using your adjuster, assuming of course that you are still in it's elastic range.
With a balance you can do a similar experiment. Place the spring on the balance pan, tare the balance and measure it's length. Apply a fixed mass to the spring, could be anything, and measure the compression. Do the calculation to find the spring constant k.
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1st January 2015, 12:46 AM #6Retired
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I must be daft, because I think my solution is still the right one.
-- You never want the sprung weight to exert down no more than 500 grams of pressure
-- You want at least, or close to, 500 grams of pressure.
--> Use the scale. Put a bit in the router, scale on the surface. Lower down until bit engages and mechanism free-floats. Unwind knob until it does not exceed 500 grams. Wind if lower than 500 to increase pressure.
Caveat: I'm assuming its a constant rate spring. I can't see, but it might be slightly conical. If not, and its a regular variable rate spring, the pressure exerted will change as the travel changes. The pressure exerted will only be true at the measured point... i.e as it becomes more and more compressed, the pressure it will exert will be greater. A constant rate spring will exert the same force across the entire non-binding range. I did a google and this seems to be the method used.... I am guessing, but I'm assuming you are engraving a surface that isn't exactly flat across its entire surface.
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1st January 2015, 04:33 PM #7Banned
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