How to draw a circle on the hud without a piece of it.
5 replies, posted
What would be the best way to draw a circle on the hud that is missing a chunk of the circle? Maybe surface.DrawCircle, if there is a way to cut out part of it after it is done? Or surface.DrawPoly which seems more reasonable but kind of expensive. I do not want to use an image, I would rather have it drawn. Does anyone have any suggestions?
When I say a circle with a piece cut out, I mean a circle that doesn't connect, kind of like a capital C.
I know of a script, but I'm not sure if I'm at liberty to give it out.
It's from a jailbreak gamemode.
Then don't reply?
That's just being dumb.
Supply a sample of what you want(paint/ps).
Is this what you want?
[lua]
if SERVER then
AddCSLuaFile()
else
function surface.DrawCirclePart(x, y, startDeg, endDeg, radius, col)
local segmentdist = (360/(math.pi*(radius)))
surface.SetDrawColor(col)
for i = startDeg, (endDeg - segmentdist) do
surface.DrawLine(x + math.cos(math.rad(i)) * radius, y - math.sin(math.rad(i)) * radius, x + math.cos(math.rad(i + segmentdist)) * radius, y - math.sin(math.rad(i + segmentdist)) * radius)
end
end
hook.Add("HUDPaint", "SurfaceCircle", function()
surface.DrawCirclePart(ScrW()/2, ScrH()/2-300, 90, 270, 100, Color(255,255,255,255))
end)
end
[/lua]
[IMG]http://puu.sh/2ULVG.jpg[/IMG]
[QUOTE=brandonj4;40667399]Is this what you want?
[lua]
if SERVER then
AddCSLuaFile()
else
function surface.DrawCirclePart(x, y, startDeg, endDeg, radius, col)
local segmentdist = (360/(math.pi*(radius)))
surface.SetDrawColor(col)
for i = startDeg, (endDeg - segmentdist) do
surface.DrawLine(x + math.cos(math.rad(i)) * radius, y - math.sin(math.rad(i)) * radius, x + math.cos(math.rad(i + segmentdist)) * radius, y - math.sin(math.rad(i + segmentdist)) * radius)
end
end
hook.Add("HUDPaint", "SurfaceCircle", function()
surface.DrawCirclePart(ScrW()/2, ScrH()/2-300, 90, 270, 100, Color(255,255,255,255))
end)
end
[/lua]
[IMG]http://puu.sh/2ULVG.jpg[/IMG][/QUOTE]
Yep! Thanks again.
Sorry, you need to Log In to post a reply to this thread.