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It is then possible to calculate the savings of a quickpause:
It is then possible to calculate the savings of a quickpause:
# previous pauses saving on this pause difference to previous
{| class="wikitable"
0 0.00s 0.00s
|+
1 0.27s 0.27s
|style="font-weight:bold"|# previous pauses||style="font-weight:bold"|saving on this pause||style="font-weight:bold"|difference to previous
2 0.47s 0.20s
|-
3 0.70s 0.23s
|0||0.00s||0.00s
4 0.90s 0.20s
|-
5 1.10s 0.20s
|1||0.27s||0.27s
6+ 1.17s 0.07s
|-
|2||0.47s||0.20s
|-
|3||0.70s||0.23s
|-
|4||0.90s||0.20s
|-
|5||1.10s||0.20s
|-
|6+||1.17s||0.07s
|}
As you can see from the 3rd column, because of rounding and little subtleties you don't get the exponential decay of savings that you'd expect.
As you can see from the 3rd column, because of rounding and little subtleties you don't get the exponential decay of savings that you'd expect.
Worked example: Silo 00A, quickpausing at the start only.
The are a total of 5 full pauses (4 plastiques and the camera).
{| class="wikitable"
|+
|style="font-weight:bold"|# QPs||style="font-weight:bold"|Saving||style="font-weight:bold"|Difference
|-
|1||1.10s||1.10s
|-
|2||2.00s||1.17 - 0.27s = 0.90s
|-
|3||2.70s||1.17 - 0.47s = 0.70s
|-
|4||3.17s||1.17 - 0.70s = 0.47s
|}
So the first 2 are certainly worthwhile, and the 3rd if it's fast enough. A 4th would probably take longer than it saved.
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