Zezo forum:

Post by zezo | 2017-10-13 | 20:04:57

Reposting this in separate thread on behalf of belharra

https://drive.google.com/open?id=0B1Ntvhh9XCUHd0VRNzVEVFVrT00

I don't have the time to maintain xls files myself, but can provide some information to someone who does, if needed.

13 Comments

Post by marcusbelgicus | 2017-10-13 | 20:08:27

Excellent source of information. I assume this is not including any hypotetical increase in speed coming from 'virtual' foils?

Post by zezo | 2017-10-13 | 20:15:54

I guess not, but we can add second formula-based sheet if need be.

Post by karriv | 2017-10-15 | 19:29:57

Great, thanks belharra and Cvetan!

Post by toxcct | 2017-10-16 | 16:05:04

How were you (or belharra) able to build the performance sheets for each sails ?

Post by zezo | 2017-10-16 | 16:49:19

It was not me, bit in the end you have 7 js arrays for the different sails and have to do linear interpolation between the data points.

In my code I do intermediate interpolation to 1 km/h / 1 degree resolution to simplify/speed up the run-time math a bit.

The general get_speed(twa,tws) function has to do two (binary) searches every time to get the interpolation data points.

Post by toxcct | 2017-10-17 | 00:49:45

Sorry, I'm a bit confused... which 'get_speed()' function are you talking about ?

I guess you're talking about your polars compiler code that is not public, right ?
(because I can't see any such function named after get_speed() in your polars.js)

Post by zezo | 2017-10-17 | 08:10:43

I don't have the exact code for the general case.

But the idea is following

Given tws, twa, arrays with twa and tws indexes and 2d polar array:

the indexes look like

"tws": [0, 2, 4, 5, 8, 10, 12, 14, 16, 18, 20, 22, 24,
"twa": [0, 20, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75,

You have to find the 4 bounding points that form square in the plane where the tws and twa are in the square

1) Find the tws index where the i-th speed is below the twa and i+1 is above
2) Same fort the twa
3) Do linear interpolation in one direction and then the other

Example: given TWA 43 and TWS 3

tws index is 1, twa index is 4 we have to interpolate between
polars[4][1], polars[4][2]
polars[5][1], polars[5][2]

The linear interpolation itself is like (for two data points, ws1 at tws[1] and ws2 at tws[2])

bs = bs1 + (bs2-bs1) * (tws - tws1)/*(tws2-tws1)

You calculate that two times for the two TWA cases and apply the same logic for the angle.

If in the example above numbers are
5 6
6 7

we get bs = 5.5 for twa 40 and 6.5 for twa 45 and then bs is 5.5 + 3/5

You can also solve the three interpolations analytically and get a single formula for the bi-linear interpolation. It looks like this

#define bilinear(x,y,A) ((A[1]+A[2]-A[0]-A[3])*x*y + (A[3]-A[2])*x + (A[0]-A[2]) * y + A[2])

where A holds the four corners and x and y are the fractional positions within the square, i.e. (tws - tws1)/*(tws2-tws1), (twa - twa1)/*(twa2-twa1)

There is additional twist above the maximum tws. Not sure if you get the last point or keep the slope defined by the last two points.

Post by karriv | 2017-10-17 | 14:14:51

There's a handy add-on available for excel for interpolation:

http://xongrid.sourceforge.net/index.htm

Post by Thistle1678 | 2017-10-19 | 21:50:58

If anyone is feeling motivated I'd like to see this in a single performance polar (always using the correct sail) I may chip away at it as I go...

Post by karriv | 2017-10-25 | 11:14:09

I compiled one single polar for the Pro-sails.

https://drive.google.com/open?id=0B3H4pv9ZBWY5cXdDUDlFa1VJMms

If you find inconsistencies in the polar, let me know.

Post by Alexandria | 2017-10-25 | 12:45:46

Thanks, karriv!

Post by Saileellely | 2017-10-25 | 11:53:48

And I have one for all sails except code 0. I used conditional formatting to indicate with colours for different sails. I also generated difference in case you do have code 0: max 3% and it is only effective between 80deg and 110deg.

I don't have google drive though for sharing.