rstudio / hex-stickers

RStudio hex stickers
Creative Commons Zero v1.0 Universal
470 stars 129 forks source link

Hex size requirements are not perfect hexagons #40

Closed schloerke closed 4 years ago

schloerke commented 4 years ago

From https://en.wikipedia.org/wiki/Hexagon#Parameters

D <- 2557
(d = sqrt(3) / 2 * D)
#> [1] 2214.427

Created on 2020-08-19 by the reprex package (v0.3.0)

This does not equal 2206px.

Reversing the formula

d <- 2206
(D = 2 / sqrt(3) * d)
#> [1] 2547.269

Created on 2020-08-19 by the reprex package (v0.3.0)

This does not equal 2557px.


I'd like to get the plumber hex added in https://github.com/rstudio/hex-stickers/pull/29 but want to be sure that I'm required to squish/stretch the image from a perfect hexagon.

cc @allisonhorst

hadley commented 4 years ago

Hmmmmm:

1.73/2 # hex.bin spec
#> [1] 0.865
2206/2557 # current recommendation
#> [1] 0.8627298
2214/2557
#> [1] 0.8658584
2206/2547
#> [1] 0.866117

Created on 2020-08-19 by the reprex package (v0.3.0)

I suspect this hasn't been a problem since the munging of the aspect ratio is so minor.

It's probably worth changing the recommendation, but if we do that I'd rather pick nicer numbers than 2206/2557 etc. Any thoughts?

schloerke commented 4 years ago

These dimensions are close to integer values when scaled.

library(magrittr)
close_integers <- function(digits) {
  (1:5000 * sqrt(3) / 2) %>% 
    round(digits = digits) %>% 
    as.character() %>% 
    grepl(".", ., fixed = TRUE) %>% 
    magrittr::not() %>% 
    which() %>% 
    setNames(., (. * sqrt(3) / 2) %>% round()) %>% 
    as.list() %>% 
    {tibble::tibble(d = as.numeric(names(.)), D = unlist(unname(.)))} %>% 
    print(n = Inf)
}
close_integers(2)
#> # A tibble: 50 x 2
#>        d     D
#>    <dbl> <int>
#>  1    84    97
#>  2   181   209
#>  3   265   306
#>  4   362   418
#>  5   446   515
#>  6   543   627
#>  7   627   724
#>  8   724   836
#>  9   808   933
#> 10   905  1045
#> 11   989  1142
#> 12  1086  1254
#> 13  1170  1351
#> 14  1254  1448
#> 15  1267  1463
#> 16  1351  1560
#> 17  1435  1657
#> 18  1532  1769
#> 19  1616  1866
#> 20  1713  1978
#> 21  1797  2075
#> 22  1894  2187
#> 23  1978  2284
#> 24  2075  2396
#> 25  2159  2493
#> 26  2256  2605
#> 27  2340  2702
#> 28  2437  2814
#> 29  2521  2911
#> 30  2605  3008
#> 31  2702  3120
#> 32  2786  3217
#> 33  2883  3329
#> 34  2967  3426
#> 35  3064  3538
#> 36  3148  3635
#> 37  3245  3747
#> 38  3329  3844
#> 39  3426  3956
#> 40  3510  4053
#> 41  3607  4165
#> 42  3691  4262
#> 43  3775  4359
#> 44  3788  4374
#> 45  3872  4471
#> 46  3956  4568
#> 47  4053  4680
#> 48  4137  4777
#> 49  4234  4889
#> 50  4318  4986
close_integers(3)
#> # A tibble: 5 x 2
#>       d     D
#>   <dbl> <int>
#> 1  1170  1351
#> 2  1351  1560
#> 3  2521  2911
#> 4  3691  4262
#> 5  3872  4471
close_integers(4)
#> # A tibble: 1 x 2
#>       d     D
#>   <dbl> <int>
#> 1  2521  2911

Created on 2020-08-19 by the reprex package (v0.3.0)

@hadley While they are not "pretty" numbers, 2521 x 2911 is a perfect match for a hexagon.

2911 * sqrt(3) / 2
#> [1] 2521

Created on 2020-08-19 by the reprex package (v0.3.0)

schloerke commented 4 years ago

There are also these pretty numbers:

close_integers(1) %>% dplyr::filter(as.character(d) %>% grepl("00$", .) | as.character(D) %>% grepl("00$", .))
#> # A tibble: 7 x 2
#>       d     D
#>   <dbl> <int>
#> 1   433   500
#> 2   866  1000
#> 3  1299  1500
#> 4  1500  1732
#> 5  1700  1963
#> 6  3200  3695
#> 7  3400  3926

close_integers(1) %>% dplyr::filter(as.character(d) %>% grepl("0$", .) & as.character(D) %>% grepl("0$", .))
#> # A tibble: 4 x 2
#>       d     D
#>   <dbl> <int>
#> 1   840   970
#> 2  1810  2090
#> 3  2650  3060
#> 4  3620  4180

2650 x 3060 isn't bad. I still put my vote in for 2521 x 2911 as it's an arbitrary number in the end, but requires no visual adjustments from the svg.

hadley commented 4 years ago

2521 x 2911 sounds good. Want to do a PR to contributing.md?

hadley commented 4 years ago

And the pull request template?

schloerke commented 4 years ago

@hadley Will do!