This certainly doesn't qualify as !!SCIENCE!! as such, because I haven't discovered anything new. But I thought it deserved posting all the same.
Earlier today I was sketching out a design for a wind farm. I was going to have 64 windmills whooshing away, and I wanted to know how much power I would lose by transfer before I got all of the power to a gear assembly that I could then lead an axle down from z0 to where it was needed in the caverns. So, I wrote a little formula to help me write this up if I ever needed to work it out again, and I thought I'd share it:
p - l = t
That more or less translates to power generated - power lost in transfer = total power left over, which sounds so obvious as to be insipid. So, in an attempt to save face, here's how it works, with explanations indented to the side with dashes to explain things in a friendly way. Keep in mind this formula only accounts for how much power is lost from the windmills to the central gear assembly: power loss from the gear assembly to whatever needs powering in your fortress is impossible for me to account for.
Definitions:
m = no. of windmills
o = no. of rows that pairs of windmills are divided into
w = wind speed (0, 20 or 40)
p = w * m -- p equals the wind speed in your area times the no. of windmills you're using
l = dependant -- the amount of power lost depends on what exactly o is
If o == 1: -- if o equals one, then l equals the no. of windmills you're using times 12
l = 12m + 5
If o > 1 and even: -- if o equals more than one and is even, then l equals the no. of windmills you're using
l = 12m + (2o + 5) times 12, plus the sum of 2 times the number of rows they're divided into + 5
If o > 1 and odd:
l = 12m + (2o + 3) -- if o equals more than one and is even, then l equals the no. of windmills you're using
-- times 12, plus the sum of 2 times the number of rows they're divided into + 3
To help clear things up, here's an example, with pretty ASCII pictures:
Key:
|
0 - windmill
|
R - Roller
= ∣∣ - Axle
☼ - Gear assembly
Windmills are at 0z
Axles, rollers and gear assemblies are at -1z
Diagram:
That there is 64 windmills in two rows, in an environment providing 40 Urists of windpower, all with the power leading to the central gear assembly which you can lead a vertical axle from downward into the earth. Here's how you use the formula to work your total power left after transfer, step by step, with dashed comments once again:
Calculation:
p - l = t
m = 64 -- we're using 64 windmills, as above
o = 2 -- these windmills are divided into two rows of pairs of windmills
w = 40 -- the wind speed is 40 (maximum speed, because we're lucky bastards)
p = 40 * 64 -- p equals our wind speed, 40, times our number of mills, 64
p = 2560 -- that comes out to 2560
l = dependant -- the amount of power lost, as we know, depends on how many rows our windmills are divided into
o > 1 & even, t/f: -- in this case, we have two rows, so o equals 2, which is more than one and is an even number
l = 12m + (2o + 5) -- the rest is just adding it all up...
l = 12m + (4 + 5)
l = 12m + 9
l = (12 * 64) + 9
l = 768 + 9
l = 777
p - l = t -- so, if we go back to our original formula
2560 - 777 = 1783 -- the amount of power generated (2560) minus the amount of power lost in transfer (777) equals
-- the amount of power we have left over at the central gear assembly, which is 1783 Urists!
I hope someone finds this useful, although I realise it's a rather niche piece of work. Also, this formula only holds up if you build your windmill batteries the same way I do: in paired groups of equal size and even numbers split about the middle. The amount of rows you divide them into isn't important: 3 rows of windmill cells with 8 windmills works out just fine, but, for example, 3 rows of 7 will not. Additionally; I'm not a mathematician. To the best of my knowledge this is correct, but YMMV. Please let me know how it works for you...if you're planning on building a windmill battery anytime soon and can be bothered, that is.
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Topic: Erm...!!MATH!! -- Windmill Power Loss/Generation Formula (Read 751 times)