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Hoping there might be some clever maths people around, that can point me in the right direction.
I'm trying to draw a 2D technical drawing for a plastic screw conveyor for glass ampoules, a bit like this one...
There's a section of constant pitch, which I have drawn, then a section of variable pitch which is 178mm long.
The pitch at the start of the variable section is 20.5mm, and at the end is 90mm, and it increases proportionally over the 178mm. The cut out for the ampoule stays the same throughout the entire screw - it is just the 'flat' bit of shaft in between the cut outs that tapers out along the variable section.
As I'm trying to draw it in 2D, I need to know where each 'flat' section starts and finishes, assuming that the centre line of the first flat section is at the start of the 178mm section.
I have a vague recollection from school of taking the equation of the variable pitch (y=mx+c) and integrating it - but it all becomes a bit vague in my head at this point!
Would appreciate any offerings.
In best Jack Sparrow voice
'That would be an ecumenical matter'
In best Jack Sparrow voice
The pirate? 🙂
Father Jack shirley?
Sorry OP can't help
Edit: I'd possibly create a cad block of the cut out and just work in from each end. One pitch at 20.5, one pitch from 90 then one in the centre 33.75 in from those
Yea not sure that works as you won't get nice smooth transitions, i see your issue
178mm doesn't seem very long as a transition from 20.5mm to 90mm. How many threads are there in the transition?
In 2D, with a variable pitch, where the flat sections are will depend on the angle the screw is rotated to.
I am thinking that your equation y = mx + c works for constant pitch.
m = pitch, x = distance moved around the tube, y = distance to flat/grooved bit from the base, depending how you want to look at it.
In that setup, with variable pitch, m needs to depend on distance moved.
m = k*x, where m is the pitch and k is the change in pitch with respect to x. (k = (90-20.5)/178).
so on the variable section y = kx^2.
For your 2D drawing, you'd need to work out the x-value for the parts of the screw that you can "see" in the drawing. You know the diameter of the screw, and so can work out the x-value for each flat bit you want to draw (N rotations around, for each of the (N + 1) flat bits you draw).
It wouldn't be unusual for me to have got completely the wrong end of the stick (screw).
For your 2D drawing, you’d need to work out the x-value for the parts of the screw that you can “see” in the drawing. You know the diameter of the screw, and so can work out the x-value for each flat bit you want to draw (N rotations around, for each of the (N + 1) flat bits you draw).
and
x = distance moved around the tube
This is where I'm struggling, because x, the distance moved around the tube, isn't a certain number of circumfrences for a given number of full turns.
The distance varies as the pitch varies.
It's beginning to screw with my mind.
I think that the x-distance is constant for a fixed number of rotations. (You're maybe thinking of adding on the y-distance as well?). The x-distance is calculated as if y=0, i.e. you remain at the base of the screw.
It is easiest to attack this problem by expressing the total angular rotation of the screw, theta, in terms of the distance x along the thread.
For a constant pitch P, this would be simple: theta = 2pi*x/P (where theta is measured in radians).
You have a varying pitch: P(x) = ((90-20.5)/178)*x + 20.5. ~ 0.39045x + 20.5
Therefore, we need to numerically integrate the angular rotation dtheta along the screw: theta = Int(2pi*dx/P(x)) between the limits 0 and 178.
We hence obtain theta ~ (2pi/0.39045)*(ln(0.39045x + 20.5)-ln(20.5))
There are 3.7889 total rotations of the screw between 0 and 178mm. Zeros (i.e. full turns) can be found at 25.078, 62.135 and 116.891mm.
I did this calculation in a rush, so you should check it carefully before using the numbers!
There are 3.7889 total rotations of the screw between 0 and 178mm. Zeros (i.e. full turns) can be found at 25.078, 62.135 and 116.891mm.
You are my Hero! and thanks Punter too, much appreciated. I've wasted about 8 hours trying to figure this out. I see an end in sight now 🙂