Hello,
I am having trouble using roundline to divide a circle into 6 equal arcs  each arc will essentially be a button. The divisions will fall where normally the shape of a star's points would be.
I am ok using Roundline to show say a volume arc that changes with a mouse action, but the basic math of this division is confusing me. My hope would be that each arc would be a separate Roundline meter completing an overall circle. So each startangle and rotationangle are what is driving me bonkers  since there is no control measure  other than what I imagine to be a calc measure but the formula per se  I ain't getting it.
Any help gisting this out would be great.
It is currently August 12th, 2020, 12:11 am
Using Roundline to divide a full circle into 6 equal arcs [solution found]

 Posts: 884
 Joined: May 7th, 2016, 7:32 am
Using Roundline to divide a full circle into 6 equal arcs [solution found]
Last edited by Mor3bane on January 15th, 2020, 1:06 pm, edited 1 time in total.
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The law of averages says what it means; even if you get everything right, you will get something wrong. Therefore; self managing error trapping initiates another set of averages  amongst the errors, some of them will not be errors, instead those instances will appear to be "luck". One cannot complain of the 'appearance' of 'infinite regress of causation', even if it does not have a predictable pattern, only that is requires luck to achieve.
There are many ways to be different  there is only one way to be yourself  be amazing at it
The law of averages says what it means; even if you get everything right, you will get something wrong. Therefore; self managing error trapping initiates another set of averages  amongst the errors, some of them will not be errors, instead those instances will appear to be "luck". One cannot complain of the 'appearance' of 'infinite regress of causation', even if it does not have a predictable pattern, only that is requires luck to achieve.

 Posts: 774
 Joined: December 30th, 2015, 9:47 am
Re: Using Roundline to divide a full circle into 6 equal arcs
Something like this?
Code: Select all
[Rainmeter]
Update=1
[Circle060]
Meter=Roundline
X=0
Y=0
W=100
H=100
StartAngle=(Rad(270))
RotationAngle=(Rad(60))
LineLength=40
LineColor=32,32,32,160
Solid=1
[Circle60120]
Meter=Roundline
X=0
Y=0
W=100
H=100
StartAngle=(Rad(330))
RotationAngle=(Rad(60))
LineLength=40
LineColor=64,64,64,160
Solid=1
[Circle120180]
Meter=Roundline
X=0
Y=0
W=100
H=100
StartAngle=(Rad(390))
RotationAngle=(Rad(60))
LineLength=40
LineColor=96,96,96,160
Solid=1
[Circle180240]
Meter=Roundline
X=0
Y=0
W=100
H=100
StartAngle=(Rad(450))
RotationAngle=(Rad(60))
LineLength=40
LineColor=128,128,128,160
Solid=1
[Circle240300]
Meter=Roundline
X=0
Y=0
W=100
H=100
StartAngle=(Rad(510))
RotationAngle=(Rad(60))
LineLength=40
LineColor=160,160,160,160
Solid=1
[Circle300360]
Meter=Roundline
X=0
Y=0
W=100
H=100
StartAngle=(Rad(570))
RotationAngle=(Rad(60))
LineLength=40
LineColor=192,192,192,160
Solid=1

 Posts: 884
 Joined: May 7th, 2016, 7:32 am
Re: Using Roundline to divide a full circle into 6 equal arcs
Beautiful!
THANKS mak_kawa!
Exactly what I needed.
I got hung up on the decimal radians rather than your formulaic way.
THANKS mak_kawa!
Exactly what I needed.
I got hung up on the decimal radians rather than your formulaic way.
My DevArt Gallery
There are many ways to be different  there is only one way to be yourself  be amazing at it
The law of averages says what it means; even if you get everything right, you will get something wrong. Therefore; self managing error trapping initiates another set of averages  amongst the errors, some of them will not be errors, instead those instances will appear to be "luck". One cannot complain of the 'appearance' of 'infinite regress of causation', even if it does not have a predictable pattern, only that is requires luck to achieve.
There are many ways to be different  there is only one way to be yourself  be amazing at it
The law of averages says what it means; even if you get everything right, you will get something wrong. Therefore; self managing error trapping initiates another set of averages  amongst the errors, some of them will not be errors, instead those instances will appear to be "luck". One cannot complain of the 'appearance' of 'infinite regress of causation', even if it does not have a predictable pattern, only that is requires luck to achieve.

 Rainmeter Sage
 Posts: 11199
 Joined: October 11th, 2010, 6:27 pm
 Location: Gheorgheni, Romania
Re: Using Roundline to divide a full circle into 6 equal arcs
Don't have to. Radians can be easily using, the only thing you have to know is that 1 PI = 180 degrees. So you can use the angles either expressed into degrees, converted to radians with a Rad() function, or expressed directly into radians and in this case you can use the PI constant:
Note that besides rewriting the angles into radians, I also decreased the values of StartAngle on the meters from [Circle120180] to [Circle300360], because the initial values have been going above 360. Definitely works this way as well, however not needed or useful, because from the point of view of the trigonometry, any angle is the same as the angle added with 360. For instance the StartAngle of the initial [Circle120180] meter is StartAngle=(Rad(390)), which is the same as StartAngle=(Rad(390360)) = StartAngle=(Rad(30)). So I reduced the angles which are larger than 360 degrees.[Circle060]
...
StartAngle=((9/2)*(PI/3))
RotationAngle=(PI/3)
;StartAngle=(Rad(270))
;RotationAngle=(Rad(60))
...
[Circle60120]
...
StartAngle=((11/2)*(PI/3))
RotationAngle=(PI/3)
;StartAngle=(Rad(330))
;RotationAngle=(Rad(60))
...
[Circle120180]
...
StartAngle=((1/2)*(PI/3))
RotationAngle=(PI/3)
;StartAngle=(Rad(30))
;RotationAngle=(Rad(60))
...
[Circle180240]
...
StartAngle=((3/2)*(PI/3))
RotationAngle=(PI/3)
;StartAngle=(Rad(90))
;RotationAngle=(Rad(60))
...
[Circle240300]
...
StartAngle=((5/2)*(PI/3))
RotationAngle=(PI/3)
;StartAngle=(Rad(150))
;RotationAngle=(Rad(60))
...
[Circle300360]
...
StartAngle=((7/2)*(PI/3))
RotationAngle=(PI/3)
;StartAngle=(Rad(210))
;RotationAngle=(Rad(60))
...

 Posts: 774
 Joined: December 30th, 2015, 9:47 am
Re: Using Roundline to divide a full circle into 6 equal arcs [solution found]
Hi Mor3bane
Sorry for bringing up solved thread.
In the Roundline meter, apparent "fill" color of a sector is in fact defined by LineColor option. That is, maybe, the outline and "fill" colors of a sector can not be drawn separately.
So, I have made a code displaying sectors with independent outline and fill colors using the Shape meter.
Possibly, this code can be written more smart...
balala, I have not used PI constant with no reason... maybe I am lazy.
Sorry for bringing up solved thread.
In the Roundline meter, apparent "fill" color of a sector is in fact defined by LineColor option. That is, maybe, the outline and "fill" colors of a sector can not be drawn separately.
So, I have made a code displaying sectors with independent outline and fill colors using the Shape meter.
Code: Select all
[Rainmeter]
Update=1
[Variables]
ArcRadius=50
[Circle060]
Meter=Shape
Shape=Path MyPath  Fill Color 32,32,32,192
MyPath=#ArcRadius#,0  ArcTo (#ArcRadius#+(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#(#ArcRadius#*Cos(Rad(60)))),#ArcRadius#  LineTo #ArcRadius#,#ArcRadius#  LineTo #ArcRadius#,0  ClosePath 1
[Circle60120]
Meter=Shape
Shape=Path MyPath  Fill Color 64,64,64,192
MyPath=(#ArcRadius#+(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#(#ArcRadius#*Cos(Rad(60))))  ArcTo (#ArcRadius#+(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#+(#ArcRadius#*Cos(Rad(60)))),#ArcRadius#  LineTo #ArcRadius#,#ArcRadius#  LineTo (#ArcRadius#+(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#(#ArcRadius#*Cos(Rad(60))))  ClosePath 1
[Circle120180]
Meter=Shape
Shape=Path MyPath  Fill Color 96,96,96,192
MyPath=(#ArcRadius#+(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#+(#ArcRadius#*Cos(Rad(60))))  ArcTo #ArcRadius#,100,#ArcRadius#  LineTo #ArcRadius#,100  LineTo #ArcRadius#,#ArcRadius#  ClosePath 1
[Circle180240]
Meter=Shape
Shape=Path MyPath  Fill Color 128,128,128,192
MyPath=#ArcRadius#,100  ArcTo (#ArcRadius#(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#+(#ArcRadius#*Cos(Rad(60)))),#ArcRadius#  LineTo #ArcRadius#,#ArcRadius#  LineTo #ArcRadius#,0  ClosePath 1
[Circle240300]
Meter=Shape
Shape=Path MyPath  Fill Color 160,160,160,192
MyPath=(#ArcRadius#(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#+(#ArcRadius#*Cos(Rad(60))))  ArcTo (#ArcRadius#(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#(#ArcRadius#*Cos(Rad(60)))),#ArcRadius#  LineTo #ArcRadius#,#ArcRadius#  LineTo (#ArcRadius#(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#+(#ArcRadius#*Cos(Rad(60))))  ClosePath 1
[Circle300360]
Meter=Shape
Shape=Path MyPath  Fill Color 192,192,192,192
MyPath=(#ArcRadius#(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#(#ArcRadius#*Cos(Rad(60))))  ArcTo #ArcRadius#,0,#ArcRadius#  LineTo #ArcRadius#,#ArcRadius#  LineTo (#ArcRadius#(#ArcRadius#*Sin(Rad(60)))),(#ArcRadius#(#ArcRadius#*Cos(Rad(60))))  ClosePath 1
balala, I have not used PI constant with no reason... maybe I am lazy.
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 Rainmeter Sage
 Posts: 11199
 Joined: October 11th, 2010, 6:27 pm
 Location: Gheorgheni, Romania
Re: Using Roundline to divide a full circle into 6 equal arcs [solution found]
No, you are not. Probably it is not obvious for everyone it can be used, however it can easier very much the work you have to do.