Sorry. Time for a snooze. Thought you were looking for ideas. Hadn’t realised it was yours. Will look closer
no problem. Always like comments. Good night
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Now I see it. Only thing I would say is to try and find a deathstar image with less in shadow. It is almost always showing just black as the shadow. I hadn’t realised it was a dual image rotation either initially.
Also, I was moving days forward with no rotation so perhaps @russellcresser can help with that.
I haven’t hit the lightbulb moment with the dual moon rotation mask thing yet.
Fantastic when the lit side comes into play though.
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Ya, I took a bit of a short cut or significant artistic license on this one. I don’t think I can find a Mimas versions that actually rotates to simulate the moon’s rotation. But also if it rotates, I’m going to lose that iconic crater that shows off it’s death star likeness so…I figure I’ll just leave it. Since the death star itself is suppose to be the “shadow”, I can’t find a completely dark version of the death star. But if it’s dark, the same characteristic dimple won’t really be visible. And if I have a completely bright version of the death star…well then it is not really a shadow…
But again, its suppose to be a joke so lots of artistic license here.
I’m hoping no one is going to rely on this to launch their personal space craft to Mimas or the death star.
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Yes . Publish a Disclaimer . Not to be used for Stella Navigation . Might lead to De-Orbiting .
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added your warning.
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You are having far too much Fun .
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Ok. Here is my next one. Not a GMT but very simple.
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Amazing first crack! Keep it up 
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Much thanks. Really appreciate feedback especially from a Partner.
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Does anyone know the shortest formula for this?
I need to test where a number falls within 3 buckets.
For example. Given a number, say less than 500, and I need to see where a number falls within 3 portions.
Portion 1 is between 1-90, Portion 2 is 91-260, and 3 is between 261-500 where 500 is the max.
The standard way of determine which portion the number sits in would be this.
=if(Number is <91, then 1, if(Number is < 261, then 2, else 3).
Is there a way in Creator to shorten this formula?
FYI, the number is random, the sizes of the portions will vary so 91 and 260 won’t be the same each time. Not hard coded.
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Facer only generates one random number on wake . For the middle test you need to test >91 && <260 && = AND .
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I do not think there would be reliable simple nested formula.
Maybe stitching 3 formulas with empty else portion would do.
$Nr<=border1?A:$$Nr>border1&&Nr<border2?B:$$Nr>=border2?C:$
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I don’t think there is a way of simplifying the structure of the ifthenelse statement. But I figure I would ask. I figure I should ask around and was hoping Creator would have something simplified.
Much thanks.
I’m actually working on correcting my GMT to factor in Daylight savings changes into my GMT. The formula is enormous.
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If the answer to the formula is really only 2 answers and not 3 separate answers but there are 3 buckets, i.e bucket one is the same answer as bucket 3, then can the equation be simplified?
If answer to Portion 1 and Portion 3 is always the same, then can if then else be simplified in creator?
Only portion 2 has a different answer. Even though there are 3 portions, there are only 2 answers…
Hope that makes sense.
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It is simple, you have to apply all 3 conditions and which ever is fulfilled, returns your desired reply. The other two return nothing. If you manage the conditions to overlap, or leave a gap, you may get two cobbled, or no answer as well.
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Use || = OR for crossing Zero .
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I dont think that is needed
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$Nr>border1&&Nr<border2?A:B$
I think this would satisfy my needs. Let me explain.
- All numbers (Nr, Border1, Border2) will be positive integers.
- There never will be a zero. If it was, then the calculation is wrong. Or the watch generated incorrect starting numbers that form the pre-calculation that derives Border1 and Border2 which is calculated way before this test. If the watch generated bad numbers to start the calculation, there is something really wrong with the watch. However, even if the numbers were really wrong, the calculation should actually still generate positive integers for Border1 and Border2. A quirk of the calculation.
- The calculation will only generate a border 1 and border 2 numbers that are always going to be around 70ish and 300ish respectively. Therefore, no zero cross over. The overall calculation for this test is not actual clock time, therefore, hours are irrelevant for this test.
- Nr will always be a positive number as it is the day number which is between 1 and 365 or 366. Therefore, no zero crossover.
Given these limitations, I don’t think the calculation would ever generate a non-answer, Nr would fall either within the borders, resulting in answer A or by default answer B.
Anyways, thank you all for helping out. I’ll run the test and see what happens. But I’m a long way off to even getting to this condition.
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