Difference between revisions of "Optimal sublayernumber for minimal friction"
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Wildly (but not totally unreasonably) assuming all factors but B are near 1: <br> | Wildly (but not totally unreasonably) assuming all factors but B are near 1: <br> | ||
| − | This means | + | |
| − | + | This means: | |
| − | That is | + | * B=32 n_crit=32768 layers. <br> |
| − | And with more layers friction losses from transport motions grow rapidly thwarting any further gains. | + | * B=10 n_crit=1000 layers. <br> |
| + | |||
| + | That is: At this point frictive losses from transport motions already make up 50% of all frictive losses. <br> | ||
| + | And with more layers friction losses from transport motions grow rapidly thwarting any further gains. <br> | ||
| + | Note that this point is already way beyond the friction loss minimum. | ||
Interestingly (and perhaps quite unexpectedly) the ratio of the bearing areas goes in as a root. | Interestingly (and perhaps quite unexpectedly) the ratio of the bearing areas goes in as a root. | ||
Revision as of 20:07, 26 August 2022
(wiki-TODO: Add a sketch to make the math much more comprehensible!)
Contents
Explanation of goals
- There is an optimal number of layers for which total friction for a given throughput T and branching factor B is minimal
- There is an number of layers for which friction from transport motion crosses and then exceeds the friction loses from assembly motions.
The aim here to derive formulas for these layer numbers respectively
- the optimal lavernumber n_opt
- the critical layernumber n_crit
Assuming constant throughput:
Why does the friction drop with increasing layer numbers for low layer numbers?
Because while the bearing area grows the operation speeds fall.
And while the area makes the friction grow linearly the speed makes the friction fall quadratically.
See: Higher throughput of smaller machinery & Math of convergent assembly
BUT the transport speed does NOT fall when adding more layers. It must stay constant.
Transport-friction grows linearly with layer numbers but more importantly
Transport friction can only be ignored so long as assembly motions are much faster than transport motions.
(TODO: Eventually make a plot identifying that friction minimum graphically.)
Math
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^3C) [/math]
We want to keep the total throughput constant while tuning the compenslow parameter to reduce frictive 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^3sFC)) [/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^3C))^2 [/math]
[math] A_A v_A^2 \gt A_T n^2 v_A^2/(B^6C^2) [/math]
[math] A_A \gt A_T n^2 /(B^6C^2) [/math]
And finally we get:
[math] n_{critical} \lt \sqrt{A_A/A_T} B^3 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
Wildly (but not totally unreasonably) assuming all factors but B are near 1:
This means:
- B=32 n_crit=32768 layers.
- B=10 n_crit=1000 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.
Note that this point is already way beyond the friction loss minimum.
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
