Difference between revisions of "Optimal sublayernumber for minimal friction"

Revision as of 07:09, 28 August 2022

(wiki-TODO: Add a sketch to make the math much more comprehensible!)

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 frictionequal layernumber n_eq

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.
• a parallel vertical stack of transport cells crossing the full height of these layers.

Speed relation from equal throughput

Total transport throughput must be equal to the
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 (assuming cube shaped chambers with equal sidelengths):
[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 C that is close to 1.
[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]
[math] v_A = v_T B^3 C / n [/math]

Discussion of preliminary result:
For one single layer n=1 assembly motions are B³C times faster than transport motions.
For n=B³C assemly motions are the same speed as transport motions.
The chamber to part size ratio F falls out of the equations.

Determining layernumber of minimal friction

Transport friction losses:
[math] P_T = \gamma (n A_T) v_T^2 [/math]

Assembly friction losses:
[math] P_A = \gamma (n A_A) (v_A)^2 [/math]
[math] P_A = \gamma (n A_A) ((v_T B^3 C)/n)^2 [/math]
[math] P_A = \gamma A_A v_T^2 B^6 C^2 / n [/math]

Total friction losses are comprised of transport & assembly losses:
[math] P_{total} = \gamma \sum_i A_i v_i^2 [/math]
[math] P_{total} = (A_T v_T^2 + A_A v_A^2) n \gamma [/math]
[math] P_{total} = \gamma v_T^2 (n A_T + A_A B^6 C^2/n) [/math]

Taking the derivative with respect to layernumber to find a the minimum for the friction losses:
[math] d P_{total} / d n = \gamma v_T^2 (A_T - A_A B^6 C^2/n_{opt}^2) = 0[/math]
[math] A_T = A_A B^6 C^2/n_{opt}^2 [/math]
And finally we get:
[math] n_{opt} = \sqrt{A_A/A_T} B^3 C [/math]

Collary:
Substituting that optimal layer into the expression for the assembly speed we get:
[math] v_{A,opt} = \sqrt{A_T/A_A} v_T [/math]

Side-note:
Transport bearing area A_T needs to cover both transport in and transport out,
since we did not consider these separately.

Determining layernumber of equal friction

No we ask for the transport losses to equal the assembly losses:
[math] A_T v_T^2 = A_A v_A^2 [/math]
Substituting v_a with our result for v_T:
[math] A_T = A_A B^6 C^2/n_{eq}^2 [/math]
And finally we get:
[math] n_{eq} = \sqrt{A_A/A_T} B^3 C [/math]

Analysis of result

Turns out this that ...

• the layermumber of equal friction coincides with
• the layernumber of minimal friction.

At this layernumber friction is minimal. At this layernumber frictive losses from transport motions make up 50% of all frictive losses.
With more layers friction losses from transport motions grow linearly thwarting any further gains.

Interestingly (and perhaps quite unexpectedly) the ratio of the bearing areas goes in as a root.

Example

Wildly (but not totally unreasonably) assuming all factors but B are near 1:

This means:

• B=32 => n_opt=32768 layers.
  
• B=10 => n_opt=1000 layers.
  

This is a very favorable/convenient result.
(TODO: Check where the proposed speeds af ~5mm/s for nanofactories lie on the P_total(n) curve.)

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

Side-note 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.

Instead of resolving for n we can also revolve for v_A:
TODO

Related

• Higher throughput of smaller machinery
• Math of convergent assembly

External links

Qualitative shape of friction-losses plotted over layer-number:
plot(10^6/n+n,n,1,5*10^3) (on wolfram alpha)