Difference between revisions of "Optimal sublayernumber for minimal friction"
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Revision as of 14:50, 26 August 2022
(wiki-TODO: Add a sketch to make the math much more comprehensible!)
Let's consider one single vertical stack of assembly cells n layers high.
Total transport throughput must be equal to total assembly throughput
[math] T_T = T_A [/math]
Total transport throughput is product-volume times time per that volume.
This time is transport-speed per chamber sidelength:
[math] T_T = V (v_T/(sF)) [/math]
Total assembly throughput is layer-number n times product-volume V per the time that it takes for assembly of that.
This time is assembly-speed per total-motion-length of assembly of one volume V.
This total-motion-length is the chamber-sidelength sF times the number-of-parts B^3 times some factor close to 1 C.
[math] T_A = n V (v_A/(B^3sFC)) [/math]
Now equating the two and simplifying we get:
[math] v_T = n v_A/(B^2C) [/math]
We want to keep the total throughput constant while tuning the compenslow parameter to reduce frictibve losses.
To that end we assume that the layer number times the assembly speed is constant. This should be intuitively clear.
Halving the speed while doubling the amount of machinery (or vice versa) the throughput stays constant.
So we assume:
[math] n v_A = X = const[/math]
Side-note: Substituting assembly the assembly-speed in the formula for the assembly throughput you get:
[math] T_A = V (X/(B^2sFC)) [/math]
This is all constants, confirming our assumption that our introduced condition will cause that.
Now let's look at the frictive losses:
[math] P = A v^2 \gamma [/math]
We split it up into assembly and transport loses:
[math] P = (A_A v_A^2 + A_T v_T^2) n \gamma [/math]
No we ask for the transport losses to stay smaller than the assembly losses:
[math] A_A v_A^2 \gt A_T v_T^2 [/math]
Substitute our result for v_T:
[math] A_A v_A^2 \gt A_T (n v_A/(B^2C))^2 [/math]
[math] A_A v_A^2 \gt A_T n^2 v_A^2/(B^4C^2) [/math]
[math] A_A \gt A_T n^2 /(B^4C^2) [/math]
And finally we get:
[math] n \lt \sqrt{A_A/A_T} B^2 C[/math]
Side-note: Transport beating area A_T needs to cover both transport in and transport out,
since we did not consider these separately.
Instead of resolving for n we can also revolve for v_A:
TODO
Analysis of result
Assuming all factors but B are near 1:
This means that for a branching factor B of e.g. B=32
"higher throughput of smaller machinery" breaks down for n=1024 layers.
That is at this point frictive losses from transport motions already make up 50% of all frictive losses.
And with more layers friction losses from transport motions grow rapidly thwarting any further gains.
Interestingly (and perhaps quite unexpectedly) the ratio of the bearing areas goes in as a root.
Legend/Key for the variables
- n … number of layers
- V … product volume
- s … product sidelength
- sF … chamber sidelength
- X … constant proportional to total throughput T
- gamma … dynamic friction coefficient
- v_A … assembly speed
- T_A … assembly throughput
- A_A … bearing area per assembly unit
- v_T … transport speed
- T_T … transport throughput
- A_T … bearing area per transport track passing one assembly unit
