Problem lining up same font with different heights.

[Pete]

So far I can't see any way to use the _U keywords.

This algorithm of .8 seems the closest for this font...

Code: (Select All)

$Color:32

Screen _NewImage(1300, 600, 32)

_ScreenMove 20, 0

Cls: _Display

Dim f&(9)

f&(1) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 16)

f&(2) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 20)

f&(3) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 24)

f&(4) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 30)

f&(5) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 38)

f&(6) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 48)

f&(7) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 58)

f&(8) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 68)

f&(9) = _LoadFont(Environ$("SYSTEMROOT") + "\Fonts\lucon.ttf", 90)

f& = 1: _Font f&(f&)

row = 200

_Font (f&(1))

fh& = _FontHeight

_PrintString (20, row - fh& * .8), "O"

_Font (f&(2))

fh& = _FontHeight

_PrintString (50, row - fh& * .8), "O"

_Font (f&(3))

fh& = _FontHeight

_PrintString (80, row - fh& * .8), "O"

_Font (f&(4))

fh& = _FontHeight

_PrintString (110, row - fh& * .8), "O"

_Font (f&(5))

fh& = _FontHeight

_PrintString (140, row - fh& * .8), "O"

_Font (f&(6))

fh& = _FontHeight

_PrintString (180, row - fh& * .8), "O"

_Font (f&(7))

fh& = _FontHeight

_PrintString (220, row - fh& * .8), "O"

_Font (f&(8))

fh& = _FontHeight

_PrintString (270, row - fh& * .8), "O"

_Font (f&(9))

fh& = _FontHeight

_PrintString (330, row - fh& * .8), "O"

Line (20, 201)-(400, 201)

_Font (f&(1))

Not prefect in all instances, but probably close enough.

Pete