Blind assistance
The $fn number is very much dependant on the size and purpose of an object, for example when creating screw threads or using cylinders as axles the cylinder really needs to be round so a high $fn.
If you are just creating a piller or a pilot hole then it can be quite low.
As an example your d=10 cylinder with a $fn=100 will produce a 100-agon ie 100 faces around a pi*d circumference for a face length of roughly 0.3mm so smooth for any visual purpose
If you are going to 3D print your model then you do not need to be much more accurate than your printer will deliver and you may well end up post processing the model ie sanding it down anyway !
Have fun
Bob
On 3 Mar 2022, at 12:30, Hendrik blindguydiy@gmail.com wrote:
Hi
Thanks it does. On the matter of $fn if I always use $fn=100 or higher will that always help in making cylinders and spheres round. Do not mind the rendering time because I will be unable to see and adjust my $fn to be correct. So looking for a safe $fn number to make most spheres and cylinders round.
Thanks
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: Bob Carter mailto:caggius@gmail.com
Sent: Thursday, 03 March 2022 14:25
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] Re: GRe: Re: Blind assistance
Terry,
You are almost right.
For center = false a cylinder stands centered around x=0, y=0 and z = +h/2 ie it stands on the centre and fully above the x,y plane.
For center = true the cylinder is centered around x=0, y=0 and z = 0 so all 3 planes dissect it in the middle of the object.
Looking at the cylinder as rendered
Plus x is to the right
Plus y is away from you (you have this reversed)
Plus z is upwards
Now this is at odds with a cube which when centre=false is always drawn fully in the positive quadrant of all 3 axes and no part of it goes negative. But when center=true is rendered like a cylinder so all 3 planes dissect it in the middle of the object.
One point of caution the precision, roundness of an object depends on the values of $fa and/or $fn such that as an example $fn=6 draws a hexagonal bar and not a cylinder - upping $fn improves the roundness but ups rendering times
Hope that helps
Bob
On 3 Mar 2022, at 11:44, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi
Ok, now I officially confused.
If center=true how does it work. For if I add 5.0 to my z then I add up with the same as what I had in the first place when I asked the question.
So let us try again.
Cylinder(d=10,h=20, center=true)
“You had interpreted centre = true as I first did (another novice), thinking ‘centre of the object’.”
Q: so where is the center then?
The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
So which one of these is correct?
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: terrypingm@gmail.com mailto:terrypingm@gmail.com
Sent: Thursday, 03 March 2022 11:36
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] GRe: Re: Blind assistance
Hi Hendrik,
If you add 5.0 to all your Z dimensions I think you’ll have it the way you want.
You had interpreted centre = true as I first did (another novice), thinking ‘centre of the object’.
--
Terry
On 3 Mar 2022, at 07:59, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi Matthias
So if I change the following is it then correct?
Cylinder(d=5, h=10, center=true)
Left bottom of cylinder at x=-2.5, y= 0, z= -5
Bottom Middle of cylinder at x=0, y=0,z=-5
Right bottom of cylinder x=2.5, y=0, z=-5
Top left of cylinder x=-2.5, y=0, z=5
Top middle of cylinder x =0, y=0,z=5
Top right of cylinder x=2.5, y=0, z=5
Back middle bottom of cylinder x=0, y=-2.5, z=-5
Front middle bottom of cylinder = x=0,y=2.5, z=-5
Back middle top of cylinder x=0, y=2.5, z=5
Front middle top of cylinder x=0, y=2.5,z=5
If I understand it now?
Thank you
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: Matthias Liffers mailto:m@tthi.as
Sent: Thursday, 03 March 2022 08:38
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] Re: Blind assistance
Hi Hendrik,
The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
Hope this helps.
Regards,
Matthias.
On Thu, 3 Mar 2022 at 14:23, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi all
I am new to openscad and need to design some parts. But the view or render do not help me since I am blind.
So for me to be able to do anything I need to understand some basic stuff so please bepatent with me.
I do understand the coordination system so what I firstly need to make 100% sure of is if I create a cylinder and make center=true is the circle vertically placed with the center of the cylinder vertically going up the y axis and the bottom of the cylinder at x0,y0,z0.
Cylinder(d=5, h=10, center=true)
So in my head I see it as follow:
Left bottom of cylinder at x=-2.5, y= 0, z= 0
Bottom Middle of cylinder at x=0, y=0,z=0
Right bottom of cylinder x=2.5, y=0, z=0
Top left of cylinder x=-2.5, y=10, z=0
Top middle of cylinder x =0, y=10,z=0
Top right of cylinder x=2.5, y=10, z=0
Back bottom of cylinder x=0, y=0, z=-2.5
Front bottom of cylinder = x=0,y=0, z=2.5
Back top of cylinder x=0, y=10, z=-2.5
Front top of cylinder x=0, y=10,z=2.5
So it will give me a upright cylinder.
Do I understand that correctly?
Thank you
Best Regards
Hendrik
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On 3/3/2022 3:12 AM, Jordan Brown wrote:
On 3/2/2022 11:58 PM, Hendrik wrote:
Cylinder(d=5, h=10, center=true)
Left bottom of cylinder at x=-2.5, y= 0, z= -5
Bottom Middle of cylinder at x=0, y=0,z=-5
Right bottom of cylinder x=2.5, y=0, z=-5
Top left of cylinder x=-2.5, y=0, z=5
Top middle of cylinder x =0, y=0,z=5
Top right of cylinder x=2.5, y=0, z=5
All of those look right.
Agree.
Back middle bottom of cylinder x=0, y=-2.5, z=-5
Front middle bottom of cylinder = x=0,y=2.5, z=-5
Positive Y is into the screen; toward the viewer is negative Y.
Agree.
Back middle top of cylinder x=0, y=2.5, z=5
Front middle top of cylinder x=0, y=2.5,z=5
You're missing a negative sign on the Y in the "front" case.
Disagree. Rather, he has the signs reversed on ALL of his y coordinates.
Sent from Mail for Windows
From: Douglas Miller
Sent: Thursday, 03 March 2022 15:50
To: discuss@lists.openscad.org
Subject: [OpenSCAD] Re: Blind assistance
On 3/3/2022 3:12 AM, Jordan Brown wrote:
On 3/2/2022 11:58 PM, Hendrik wrote:
Cylinder(d=5, h=10, center=true)
Left bottom of cylinder at x=-2.5, y= 0, z= -5
Bottom Middle of cylinder at x=0, y=0,z=-5
Right bottom of cylinder x=2.5, y=0, z=-5
Top left of cylinder x=-2.5, y=0, z=5
Top middle of cylinder x =0, y=0,z=5
Top right of cylinder x=2.5, y=0, z=5
All of those look right.
Agree.
Back middle bottom of cylinder x=0, y=-2.5, z=-5
Front middle bottom of cylinder = x=0,y=2.5, z=-5
Positive Y is into the screen; toward the viewer is negative Y.
Agree.
Back middle top of cylinder x=0, y=2.5, z=5
Front middle top of cylinder x=0, y=2.5,z=5
You're missing a negative sign on the Y in the "front" case.
Disagree. Rather, he has the signs reversed on ALL of his y coordinates.
Is there any way to work out on the size of the object what the $fn or $fs and $fa must be?
Sent from Mail for Windows
From: Bob Carter
Sent: Thursday, 03 March 2022 15:36
To: OpenSCAD general discussion
Subject: [OpenSCAD] Re: GRe: Re: Blind assistance
The $fn number is very much dependant on the size and purpose of an object, for example when creating screw threads or using cylinders as axles the cylinder really needs to be round so a high $fn.
If you are just creating a piller or a pilot hole then it can be quite low.
As an example your d=10 cylinder with a $fn=100 will produce a 100-agon ie 100 faces around a pi*d circumference for a face length of roughly 0.3mm so smooth for any visual purpose
If you are going to 3D print your model then you do not need to be much more accurate than your printer will deliver and you may well end up post processing the model ie sanding it down anyway !
Have fun
Bob
On 3 Mar 2022, at 12:30, Hendrik <blindguydiy@gmail.com> wrote:
Hi
Thanks it does. On the matter of $fn if I always use $fn=100 or higher will that always help in making cylinders and spheres round. Do not mind the rendering time because I will be unable to see and adjust my $fn to be correct. So looking for a safe $fn number to make most spheres and cylinders round.
Thanks
Sent from Mail for Windows
From: Bob Carter
Sent: Thursday, 03 March 2022 14:25
To: OpenSCAD general discussion
Subject: [OpenSCAD] Re: GRe: Re: Blind assistanceTerry,
You are almost right.
For center = false a cylinder stands centered around x=0, y=0 and z = +h/2 ie it stands on the centre and fully above the x,y plane.
For center = true the cylinder is centered around x=0, y=0 and z = 0 so all 3 planes dissect it in the middle of the object.
Looking at the cylinder as rendered
Plus x is to the right
Plus y is away from you (you have this reversed)
Plus z is upwards
Now this is at odds with a cube which when centre=false is always drawn fully in the positive quadrant of all 3 axes and no part of it goes negative. But when center=true is rendered like a cylinder so all 3 planes dissect it in the middle of the object.
One point of caution the precision, roundness of an object depends on the values of $fa and/or $fn such that as an example $fn=6 draws a hexagonal bar and not a cylinder - upping $fn improves the roundness but ups rendering times
Hope that helps
Bob
On 3 Mar 2022, at 11:44, Hendrik <blindguydiy@gmail.com> wrote:
Hi
Ok, now I officially confused.
If center=true how does it work. For if I add 5.0 to my z then I add up with the same as what I had in the first place when I asked the question.
So let us try again.
Cylinder(d=10,h=20, center=true)
“You had interpreted centre = true as I first did (another novice), thinking ‘centre of the object’.”
Q: so where is the center then?
The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
So which one of these is correct?
Sent from Mail for Windows
From: terrypingm@gmail.com
Sent: Thursday, 03 March 2022 11:36
To: OpenSCAD general discussion
Subject: [OpenSCAD] GRe: Re: Blind assistanceHi Hendrik,
If you add 5.0 to all your Z dimensions I think you’ll have it the way you want.
You had interpreted centre = true as I first did (another novice), thinking ‘centre of the object’.
--
Terry
On 3 Mar 2022, at 07:59, Hendrik <blindguydiy@gmail.com> wrote:
Hi Matthias
So if I change the following is it then correct?
Cylinder(d=5, h=10, center=true)
Left bottom of cylinder at x=-2.5, y= 0, z= -5
Bottom Middle of cylinder at x=0, y=0,z=-5
Right bottom of cylinder x=2.5, y=0, z=-5
Top left of cylinder x=-2.5, y=0, z=5
Top middle of cylinder x =0, y=0,z=5
Top right of cylinder x=2.5, y=0, z=5
Back middle bottom of cylinder x=0, y=-2.5, z=-5
Front middle bottom of cylinder = x=0,y=2.5, z=-5
Back middle top of cylinder x=0, y=2.5, z=5
Front middle top of cylinder x=0, y=2.5,z=5
If I understand it now?
Thank you
Sent from Mail for Windows
From: Matthias Liffers
Sent: Thursday, 03 March 2022 08:38
To: OpenSCAD general discussion
Subject: [OpenSCAD] Re: Blind assistanceHi Hendrik,
The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
Hope this helps.
Regards,
Matthias.
On Thu, 3 Mar 2022 at 14:23, Hendrik <blindguydiy@gmail.com> wrote:
Hi all
I am new to openscad and need to design some parts. But the view or render do not help me since I am blind.
So for me to be able to do anything I need to understand some basic stuff so please bepatent with me.
I do understand the coordination system so what I firstly need to make 100% sure of is if I create a cylinder and make center=true is the circle vertically placed with the center of the cylinder vertically going up the y axis and the bottom of the cylinder at x0,y0,z0.
Cylinder(d=5, h=10, center=true)
So in my head I see it as follow:
Left bottom of cylinder at x=-2.5, y= 0, z= 0
Bottom Middle of cylinder at x=0, y=0,z=0
Right bottom of cylinder x=2.5, y=0, z=0
Top left of cylinder x=-2.5, y=10, z=0
Top middle of cylinder x =0, y=10,z=0
Top right of cylinder x=2.5, y=10, z=0
Back bottom of cylinder x=0, y=0, z=-2.5
Front bottom of cylinder = x=0,y=0, z=2.5
Back top of cylinder x=0, y=10, z=-2.5
Front top of cylinder x=0, y=10,z=2.5
So it will give me a upright cylinder.
Do I understand that correctly?
Thank you
Best Regards
Hendrik
Sent from Mail for Windows
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On 3/3/2022 4:34 AM, terrypingm@gmail.com wrote:
Hi Hendrik,
If you add 5.0 to all your Z dimensions I think you’ll have it the way
you want.
That, of course, depends on what he wants.
You had interpreted centre = true as I first did (another novice),
thinking ‘centre of the object’.
'center of the object' IS the correct interpretation of "center = true"
-- specifically, "center = true" places the geometric center of the
object at the origin in 3-space (0,0,0).
On 3/3/2022 7:20 AM, Hendrik wrote:
Is there any way to work out on the size of the object what the $fn or
$fs and $fa must be?
I think it's best if you use $fn only if you're deliberately trying to
create an N-gon, and otherwise use $fs and $fa.
$fs and $fa naturally adjust for the size of the object. I usually set
$fa to 10 or so, which creates an object that is visually pretty round
at large sizes, and $fs to 0.5 or so which avoids creating small objects
with ridiculously small faces.
$fa limits the minimum angle that a chord crosses. $fa=10 will suggest
a 36-gon, which is pretty round.
$fs limits the minimum length of a chord. Assuming conventional
scaling, where each unit is a millimeter, $fa=0.5 will mean that your
minimum chord size will be half a millimeter, probably down around your
extrusion width and only a few times your horizontal resolution.
Starting small, $fs will dominate until you get up to a size where the
angle across $fs is equal to $fa, and then $fa will dominate. That's at
r = $fs / (2 * sin( $fa / 2 )). For $fa=10 and $fs=0.5, that's at r=2.9.
If objects larger than that are too un-round, make $fa smaller. If
objects smaller than that are too un-round, make $fs smaller.
Another factor that may be relevant is that, at least for me, I
sometimes get extrusion artifacts at places where the head changes
direction - it's a little thicker at those points. I don't think the
combination of the slicer and the printer firmware are really getting
the extrusion to match the head's acceleration. That's hard to analyze
geometrically, but might suggest smaller values (and so more sides) than
geometry alone would suggest.
Assuming that you're 3D printing, you might want to make a few test
objects that you can feel.
Terry I am no expert but happy to answer questions if I can.
I have not used $fa and $fs that much so I can not really comment on their usage.
Simplistically the size of a facet is pi * diameter / $fn so for 10mm diameter and $fn = 100 gives a 0.3mm facet which is not noticeable but for a 100mm diameter then you would get a 3mm facet size which for me is not acceptable so I need a much higher $fn.
In the end it all comes down to your proposed use and what is acceptable for that application. You can set a default $fn globally, then on a can by case basis decide how big a facet your application can accept then reverse the calculation to set local $fn as part of each cylinder invocation. That way you need only change the ones that matter.
I am mostly printing cockpit instruments that have large diameter dial fascias and screw on bezels so I tend to use a very high $fn and a lot of slop on the screw threads.
Typically I will use a global define
$fn = $preview ? 32 : 360;
This gives me the quick F5 previews to develop with and long F6 renders for the finished work.
But I have been directed by others here to use $fa for some of my larger curves !
regards
Bob
On 3 Mar 2022, at 15:20, Hendrik blindguydiy@gmail.com wrote:
Hi
Is there any way to work out on the size of the object what the $fn or $fs and $fa must be?
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: Bob Carter mailto:caggius@gmail.com
Sent: Thursday, 03 March 2022 15:36
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] Re: GRe: Re: Blind assistance
The $fn number is very much dependant on the size and purpose of an object, for example when creating screw threads or using cylinders as axles the cylinder really needs to be round so a high $fn.
If you are just creating a piller or a pilot hole then it can be quite low.
As an example your d=10 cylinder with a $fn=100 will produce a 100-agon ie 100 faces around a pi*d circumference for a face length of roughly 0.3mm so smooth for any visual purpose
If you are going to 3D print your model then you do not need to be much more accurate than your printer will deliver and you may well end up post processing the model ie sanding it down anyway !
Have fun
Bob
On 3 Mar 2022, at 12:30, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi
Thanks it does. On the matter of $fn if I always use $fn=100 or higher will that always help in making cylinders and spheres round. Do not mind the rendering time because I will be unable to see and adjust my $fn to be correct. So looking for a safe $fn number to make most spheres and cylinders round.
Thanks
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: Bob Carter mailto:caggius@gmail.com
Sent: Thursday, 03 March 2022 14:25
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] Re: GRe: Re: Blind assistance
Terry,
You are almost right.
For center = false a cylinder stands centered around x=0, y=0 and z = +h/2 ie it stands on the centre and fully above the x,y plane.
For center = true the cylinder is centered around x=0, y=0 and z = 0 so all 3 planes dissect it in the middle of the object.
Looking at the cylinder as rendered
Plus x is to the right
Plus y is away from you (you have this reversed)
Plus z is upwards
Now this is at odds with a cube which when centre=false is always drawn fully in the positive quadrant of all 3 axes and no part of it goes negative. But when center=true is rendered like a cylinder so all 3 planes dissect it in the middle of the object.
One point of caution the precision, roundness of an object depends on the values of $fa and/or $fn such that as an example $fn=6 draws a hexagonal bar and not a cylinder - upping $fn improves the roundness but ups rendering times
Hope that helps
Bob
On 3 Mar 2022, at 11:44, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi
Ok, now I officially confused.
If center=true how does it work. For if I add 5.0 to my z then I add up with the same as what I had in the first place when I asked the question.
So let us try again.
Cylinder(d=10,h=20, center=true)
“You had interpreted centre = true as I first did (another novice), thinking ‘centre of the object’.”
Q: so where is the center then?
The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
So which one of these is correct?
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: terrypingm@gmail.com mailto:terrypingm@gmail.com
Sent: Thursday, 03 March 2022 11:36
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] GRe: Re: Blind assistance
Hi Hendrik,
If you add 5.0 to all your Z dimensions I think you’ll have it the way you want.
You had interpreted centre = true as I first did (another novice), thinking ‘centre of the object’.
--
Terry
On 3 Mar 2022, at 07:59, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi Matthias
So if I change the following is it then correct?
Cylinder(d=5, h=10, center=true)
Left bottom of cylinder at x=-2.5, y= 0, z= -5
Bottom Middle of cylinder at x=0, y=0,z=-5
Right bottom of cylinder x=2.5, y=0, z=-5
Top left of cylinder x=-2.5, y=0, z=5
Top middle of cylinder x =0, y=0,z=5
Top right of cylinder x=2.5, y=0, z=5
Back middle bottom of cylinder x=0, y=-2.5, z=-5
Front middle bottom of cylinder = x=0,y=2.5, z=-5
Back middle top of cylinder x=0, y=2.5, z=5
Front middle top of cylinder x=0, y=2.5,z=5
If I understand it now?
Thank you
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: Matthias Liffers mailto:m@tthi.as
Sent: Thursday, 03 March 2022 08:38
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] Re: Blind assistance
Hi Hendrik,
The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
Hope this helps.
Regards,
Matthias.
On Thu, 3 Mar 2022 at 14:23, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi all
I am new to openscad and need to design some parts. But the view or render do not help me since I am blind.
So for me to be able to do anything I need to understand some basic stuff so please bepatent with me.
I do understand the coordination system so what I firstly need to make 100% sure of is if I create a cylinder and make center=true is the circle vertically placed with the center of the cylinder vertically going up the y axis and the bottom of the cylinder at x0,y0,z0.
Cylinder(d=5, h=10, center=true)
So in my head I see it as follow:
Left bottom of cylinder at x=-2.5, y= 0, z= 0
Bottom Middle of cylinder at x=0, y=0,z=0
Right bottom of cylinder x=2.5, y=0, z=0
Top left of cylinder x=-2.5, y=10, z=0
Top middle of cylinder x =0, y=10,z=0
Top right of cylinder x=2.5, y=10, z=0
Back bottom of cylinder x=0, y=0, z=-2.5
Front bottom of cylinder = x=0,y=0, z=2.5
Back top of cylinder x=0, y=10, z=-2.5
Front top of cylinder x=0, y=10,z=2.5
So it will give me a upright cylinder.
Do I understand that correctly?
Thank you
Best Regards
Hendrik
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
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I only use $fn when I want a specific number of sides, for example to make
a hexagon. My global definitions only set $fa and $fs suitable for 3D
printing. I use $fa = 6 and $fs = half my extrusion width, which is usually
0.25mm. When milling I set $fa = 1 because it makes sharper corners that
are more noticeable.
On Thu, 3 Mar 2022 at 18:51, Bob Carter caggius@gmail.com wrote:
Terry I am no expert but happy to answer questions if I can.
I have not used $fa and $fs that much so I can not really comment on their
usage.
Simplistically the size of a facet is pi * diameter / $fn so for 10mm
diameter and $fn = 100 gives a 0.3mm facet which is not noticeable but for
a 100mm diameter then you would get a 3mm facet size which for me is not
acceptable so I need a much higher $fn.
In the end it all comes down to your proposed use and what is acceptable
for that application. You can set a default $fn globally, then on a can
by case basis decide how big a facet your application can accept then
reverse the calculation to set local $fn as part of each cylinder
invocation. That way you need only change the ones that matter.
I am mostly printing cockpit instruments that have large diameter dial
fascias and screw on bezels so I tend to use a very high $fn and a lot of
slop on the screw threads.
Typically I will use a global define
$fn = $preview ? 32 : 360;
This gives me the quick F5 previews to develop with and long F6 renders
for the finished work.
But I have been directed by others here to use $fa for some of my larger
curves !
regards
Bob
On 3 Mar 2022, at 15:20, Hendrik blindguydiy@gmail.com wrote:
Hi
Is there any way to work out on the size of the object what the $fn or $fs
and $fa must be?
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for
Windows
*From: *Bob Carter caggius@gmail.com
*Sent: *Thursday, 03 March 2022 15:36
*To: *OpenSCAD general discussion discuss@lists.openscad.org
*Subject: *[OpenSCAD] Re: GRe: Re: Blind assistance
The $fn number is very much dependant on the size and purpose of an
object, for example when creating screw threads or using cylinders as axles
the cylinder really needs to be round so a high $fn.
If you are just creating a piller or a pilot hole then it can be quite
low.
As an example your d=10 cylinder with a $fn=100 will produce a 100-agon ie
100 faces around a pi*d circumference for a face length of roughly 0.3mm so
smooth for any visual purpose
If you are going to 3D print your model then you do not need to be much
more accurate than your printer will deliver and you may well end up post
processing the model ie sanding it down anyway !
Have fun
Bob
On 3 Mar 2022, at 12:30, Hendrik blindguydiy@gmail.com wrote:
Hi
Thanks it does. On the matter of $fn if I always use $fn=100 or higher
will that always help in making cylinders and spheres round. Do not mind
the rendering time because I will be unable to see and adjust my $fn to be
correct. So looking for a safe $fn number to make most spheres and
cylinders round.
Thanks
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for
Windows
*From: *Bob Carter caggius@gmail.com
*Sent: *Thursday, 03 March 2022 14:25
*To: *OpenSCAD general discussion discuss@lists.openscad.org
*Subject: *[OpenSCAD] Re: GRe: Re: Blind assistance
Terry,
You are almost right.
For center = false a cylinder stands centered around x=0, y=0 and z = +h/2
ie it stands on the centre and fully above the x,y plane.
For center = true the cylinder is centered around x=0, y=0 and z = 0 so
all 3 planes dissect it in the middle of the object.
Looking at the cylinder as rendered
Plus x is to the right
Plus y is away from you (you have this reversed)
Plus z is upwards
Now this is at odds with a cube which when centre=false is always drawn
fully in the positive quadrant of all 3 axes and no part of it goes
negative. But when center=true is rendered like a cylinder so all 3 planes
dissect it in the middle of the object.
One point of caution the precision, roundness of an object depends on the
values of $fa and/or $fn such that as an example $fn=6 draws a hexagonal
bar and not a cylinder - upping $fn improves the roundness but ups
rendering times
Hope that helps
Bob
On 3 Mar 2022, at 11:44, Hendrik blindguydiy@gmail.com wrote:
Hi
Ok, now I officially confused.
If center=true how does it work. For if I add 5.0 to my z then I add up
with the same as what I had in the first place when I asked the question.
So let us try again.
Cylinder(d=10,h=20, center=true)
“You had interpreted centre = true as I first did (another novice),
thinking ‘centre of the object’.”
Q: so where is the center then?
The rendered cylinder will be upright, with its very centre at the origin
(0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its
centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal
circle of diameter 5 with its centre at x=0, y=0, z=-5.
So which one of these is correct?
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for
Windows
*From: *terrypingm@gmail.com
*Sent: *Thursday, 03 March 2022 11:36
*To: *OpenSCAD general discussion discuss@lists.openscad.org
*Subject: *[OpenSCAD] GRe: Re: Blind assistance
Hi Hendrik,
If you add 5.0 to all your Z dimensions I think you’ll have it the way you
want.
You had interpreted centre = true as I first did (another novice),
thinking ‘centre of the object’.
--
Terry
On 3 Mar 2022, at 07:59, Hendrik blindguydiy@gmail.com wrote:
Hi Matthias
So if I change the following is it then correct?
Cylinder(d=5, h=10, center=true)
Left bottom of cylinder at x=-2.5, y= 0, z= -5
Bottom Middle of cylinder at x=0, y=0,z=-5
Right bottom of cylinder x=2.5, y=0, z=-5
Top left of cylinder x=-2.5, y=0, z=5
Top middle of cylinder x =0, y=0,z=5
Top right of cylinder x=2.5, y=0, z=5
Back middle bottom of cylinder x=0, y=-2.5, z=-5
Front middle bottom of cylinder = x=0,y=2.5, z=-5
Back middle top of cylinder x=0, y=2.5, z=5
Front middle top of cylinder x=0, y=2.5,z=5
If I understand it now?
Thank you
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for
Windows
*From: *Matthias Liffers m@tthi.as
*Sent: *Thursday, 03 March 2022 08:38
*To: *OpenSCAD general discussion discuss@lists.openscad.org
*Subject: *[OpenSCAD] Re: Blind assistance
Hi Hendrik,
The rendered cylinder will be upright, with its very centre at the origin
(0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its
centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal
circle of diameter 5 with its centre at x=0, y=0, z=-5.
Hope this helps.
Regards,
Matthias.
On Thu, 3 Mar 2022 at 14:23, Hendrik blindguydiy@gmail.com wrote:
Hi all
I am new to openscad and need to design some parts. But the view or
render do not help me since I am blind.
So for me to be able to do anything I need to understand some basic stuff
so please bepatent with me.
I do understand the coordination system so what I firstly need to make
100% sure of is if I create a cylinder and make center=true is the circle
vertically placed with the center of the cylinder vertically going up the y
axis and the bottom of the cylinder at x0,y0,z0.
Cylinder(d=5, h=10, center=true)
So in my head I see it as follow:
Left bottom of cylinder at x=-2.5, y= 0, z= 0
Bottom Middle of cylinder at x=0, y=0,z=0
Right bottom of cylinder x=2.5, y=0, z=0
Top left of cylinder x=-2.5, y=10, z=0
Top middle of cylinder x =0, y=10,z=0
Top right of cylinder x=2.5, y=10, z=0
Back bottom of cylinder x=0, y=0, z=-2.5
Front bottom of cylinder = x=0,y=0, z=2.5
Back top of cylinder x=0, y=10, z=-2.5
Front top of cylinder x=0, y=10,z=2.5
So it will give me a upright cylinder.
Do I understand that correctly?
Thank you
Best Regards
Hendrik
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Windows
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Hi Bob, Hi Hendrick,
Sorry if I muddied the water! I was not only wrong about the syntax but also
potentially added confusion by typing 'z dimensions' instead of 'z coordinates'.
Bob: Here and in your later post, are you replying to me, or to Hendrick?
Hendrick: Have you now got what you needed? Also, would it help you if replies
attached an audio narrative file?
Terry
====================
On Thu, 3 Mar 2022 12:24:35 +0000, you wrote:
Terry,
You are almost right.
For center = false a cylinder stands centered around x=0, y=0 and z = +h/2 ie it stands on the centre and fully above the x,y plane.
For center = true the cylinder is centered around x=0, y=0 and z = 0 so all 3 planes dissect it in the middle of the object.
Looking at the cylinder as rendered
Plus x is to the right
Plus y is away from you (you have this reversed)
Plus z is upwards
Now this is at odds with a cube which when centre=false is always drawn fully in the positive quadrant of all 3 axes and no part of it goes negative. But when center=true is rendered like a cylinder so all 3 planes dissect it in the middle of the object.
One point of caution the precision, roundness of an object depends on the values of $fa and/or $fn such that as an example $fn=6 draws a hexagonal bar and not a cylinder - upping $fn improves the roundness but ups rendering times
Hope that helps
Bob
On 3 Mar 2022, at 11:44, Hendrik blindguydiy@gmail.com wrote:
Hi
Ok, now I officially confused.
If center=true how does it work. For if I add 5.0 to my z then I add up with the same as what I had in the first place when I asked the question.
So let us try again.
Cylinder(d=10,h=20, center=true)
You had interpreted centre = true as I first did (another novice), thinking centre of the object.
Q: so where is the center then?The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
So which one of these is correct?
Sent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: terrypingm@gmail.com mailto:terrypingm@gmail.com
Sent: Thursday, 03 March 2022 11:36
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] GRe: Re: Blind assistanceHi Hendrik,
If you add 5.0 to all your Z dimensions I think youll have it the way you want.
You had interpreted centre = true as I first did (another novice), thinking centre of the object.
--
TerryOn 3 Mar 2022, at 07:59, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
?
Hi MatthiasSo if I change the following is it then correct?
Cylinder(d=5, h=10, center=true)
Left bottom of cylinder at x=-2.5, y= 0, z= -5
Bottom Middle of cylinder at x=0, y=0,z=-5
Right bottom of cylinder x=2.5, y=0, z=-5
Top left of cylinder x=-2.5, y=0, z=5
Top middle of cylinder x =0, y=0,z=5
Top right of cylinder x=2.5, y=0, z=5
Back middle bottom of cylinder x=0, y=-2.5, z=-5
Front middle bottom of cylinder = x=0,y=2.5, z=-5
Back middle top of cylinder x=0, y=2.5, z=5
Front middle top of cylinder x=0, y=2.5,z=5If I understand it now?
Thank youSent from Mail https://go.microsoft.com/fwlink/?LinkId=550986 for Windows
From: Matthias Liffers mailto:m@tthi.as
Sent: Thursday, 03 March 2022 08:38
To: OpenSCAD general discussion mailto:discuss@lists.openscad.org
Subject: [OpenSCAD] Re: Blind assistanceHi Hendrik,
The rendered cylinder will be upright, with its very centre at the origin (0,0,0).
The top of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=5 and the bottom of the cylinder is a horizontal circle of diameter 5 with its centre at x=0, y=0, z=-5.
Hope this helps.
Regards,
Matthias.
On Thu, 3 Mar 2022 at 14:23, Hendrik <blindguydiy@gmail.com mailto:blindguydiy@gmail.com> wrote:
Hi allI am new to openscad and need to design some parts. But the view or render do not help me since I am blind.
So for me to be able to do anything I need to understand some basic stuff so please bepatent with me.
I do understand the coordination system so what I firstly need to make 100% sure of is if I create a cylinder and make center=true is the circle vertically placed with the center of the cylinder vertically going up the y axis and the bottom of the cylinder at x0,y0,z0.
Cylinder(d=5, h=10, center=true)
So in my head I see it as follow:
Left bottom of cylinder at x=-2.5, y= 0, z= 0
Bottom Middle of cylinder at x=0, y=0,z=0
Right bottom of cylinder x=2.5, y=0, z=0
Top left of cylinder x=-2.5, y=10, z=0
Top middle of cylinder x =0, y=10,z=0
Top right of cylinder x=2.5, y=10, z=0
Back bottom of cylinder x=0, y=0, z=-2.5
Front bottom of cylinder = x=0,y=0, z=2.5
Back top of cylinder x=0, y=10, z=-2.5
Front top of cylinder x=0, y=10,z=2.5So it will give me a upright cylinder.
Do I understand that correctly?
Thank you
Best Regards
Hendrik
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