Some trigonometric issue
Hello,
I have a rectangle abcd with known width w and height h.
The rectangle is rotated according to the theta angle and becomes ab'c'd'.
I can calculate the coordinates of the rotated rectangle. However my
knowledge in trigonometry is too weak to find the position of the point p on
the y axis. This point is at the intersection of the y-axis and the line
passing through the upper side of the rectangle after rotation.
Does anyone know how to calculate this? The distance between a and b' can be
easily calculated, but from there I am not quite sure how to proceed.
http://forum.openscad.org/file/t2715/trigonometry_question.jpg
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I think it is just length(ab) / acos(theta)
On Fri, 16 Oct 2020 at 11:29, amundsen roald.baudoux@brutele.be wrote:
Hello,
I have a rectangle abcd with known width w and height h.
The rectangle is rotated according to the theta angle and becomes ab'c'd'.
I can calculate the coordinates of the rotated rectangle. However my
knowledge in trigonometry is too weak to find the position of the point p
on
the y axis. This point is at the intersection of the y-axis and the line
passing through the upper side of the rectangle after rotation.
Does anyone know how to calculate this? The distance between a and b' can
be
easily calculated, but from there I am not quite sure how to proceed.
http://forum.openscad.org/file/t2715/trigonometry_question.jpg
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On Fri, Oct 16, 2020 at 11:34:41AM +0100, nop head wrote:
I think it is just length(ab) / acos(theta)
Rotate the "whole figure" back along theta, but take the Y-axis along.
So now you want to find the intersection of the rotated Y-axis
and the upper side of your rectangle.
The line is: y/x = tan (theta).
and the topside of the rectangle is y = h
So I think you get h = tan (theta) * x or x = h/tan (theta)
So x,h is the intersection point, and now you calculate the lenght from
the origin to find the Y-coordinate requested in the original question.
yp = sqrt (hh + h/tan (theta) h/tan (theta))
You could/should check that x = h/tan (theta) is less than
w. Otherwise you're intersecting with the other edge of the rectangle.
But I'm guessing you have good reasons to believe that this will never
happen.
Roger.
On Fri, 16 Oct 2020 at 11:29, amundsen roald.baudoux@brutele.be wrote:
Hello,
I have a rectangle abcd with known width w and height h.
The rectangle is rotated according to the theta angle and becomes ab'c'd'.
I can calculate the coordinates of the rotated rectangle. However my
knowledge in trigonometry is too weak to find the position of the point p
on
the y axis. This point is at the intersection of the y-axis and the line
passing through the upper side of the rectangle after rotation.
Does anyone know how to calculate this? The distance between a and b' can
be
easily calculated, but from there I am not quite sure how to proceed.
http://forum.openscad.org/file/t2715/trigonometry_question.jpg
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** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110 **
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
The triangle pab' is a right angle triangle and the angle pab' is equal to
theta. So ab' / p = cos(theta). Hence p = ab' / cos(theta) and ab' = ab. So
ab / cos(theta).
On Fri, 16 Oct 2020 at 11:53, Rogier Wolff R.E.Wolff@bitwizard.nl wrote:
On Fri, Oct 16, 2020 at 11:34:41AM +0100, nop head wrote:
I think it is just length(ab) / acos(theta)
Rotate the "whole figure" back along theta, but take the Y-axis along.
So now you want to find the intersection of the rotated Y-axis
and the upper side of your rectangle.
The line is: y/x = tan (theta).
and the topside of the rectangle is y = h
So I think you get h = tan (theta) * x or x = h/tan (theta)
So x,h is the intersection point, and now you calculate the lenght from
the origin to find the Y-coordinate requested in the original question.
yp = sqrt (hh + h/tan (theta) h/tan (theta))
You could/should check that x = h/tan (theta) is less than
w. Otherwise you're intersecting with the other edge of the rectangle.
But I'm guessing you have good reasons to believe that this will never
happen.
Roger.
On Fri, 16 Oct 2020 at 11:29, amundsen roald.baudoux@brutele.be wrote:
Hello,
I have a rectangle abcd with known width w and height h.
The rectangle is rotated according to the theta angle and becomes
ab'c'd'.
I can calculate the coordinates of the rotated rectangle. However my
knowledge in trigonometry is too weak to find the position of the
point p
on
the y axis. This point is at the intersection of the y-axis and the
line
passing through the upper side of the rectangle after rotation.
Does anyone know how to calculate this? The distance between a and b'
can
be
easily calculated, but from there I am not quite sure how to proceed.
http://forum.openscad.org/file/t2715/trigonometry_question.jpg
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** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110
**
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
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On 2020-10-16 12:28, amundsen wrote:
I can calculate the coordinates of the rotated rectangle. However my
knowledge in trigonometry is too weak to find the position of the point
p on
the y axis.
A general solution is based on 2 points on each line + using
determinants, details at:
https://en.wikipedia.org/wiki/Line%E2%80%93line_intersection#Given_two_points_on_each_line
Se the formula for (Px,Py)
Regards
Carsten Arholm
ab/cos(theta) seems to give the right position.
Thank you everyone!
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On Fri, Oct 16, 2020 at 12:04:26PM +0100, nop head wrote:
The triangle pab' is a right angle triangle and the angle pab' is equal to
theta. So ab' / p = cos(theta). Hence p = ab' / cos(theta) and ab' = ab. So
ab / cos(theta).
Although I might get a different formula... If I did things right the
result should be the same. It is possible that without pen-and-paper
I've done something wrong. And my trig is a bit rusty.
If you figure out a formula and want to double check it you can put in
the numbers that you can verify.
If my rectangle is 2 wide, 1 high and theta is 45deg. then the answer
should be sqrt (2).
if a is the width and b is the height, then ab/cos(theta) becomes 1.41
precisely the answer we expect....
But if our rectangle is 3 wide and the height and theta still 1 and 45
then the answer should still be 1.41 (sqrt(2)), but your formula gives
something else. The first example was a coincidence!
My formula however comes to yp = h * sqrt(1+1/tan(theta)^2) = sqrt(2)
for both examples. We can try a few more "known examples", but so far
my formula seems to do a bit better than yours.
Roger.
P.S. I started out thinking that both formulas would come out to the
same number but it seems that this is not the case.
On Fri, 16 Oct 2020 at 11:53, Rogier Wolff R.E.Wolff@bitwizard.nl wrote:
On Fri, Oct 16, 2020 at 11:34:41AM +0100, nop head wrote:
I think it is just length(ab) / acos(theta)
Rotate the "whole figure" back along theta, but take the Y-axis along.
So now you want to find the intersection of the rotated Y-axis
and the upper side of your rectangle.
The line is: y/x = tan (theta).
and the topside of the rectangle is y = h
So I think you get h = tan (theta) * x or x = h/tan (theta)
So x,h is the intersection point, and now you calculate the lenght from
the origin to find the Y-coordinate requested in the original question.
yp = sqrt (hh + h/tan (theta) h/tan (theta))
You could/should check that x = h/tan (theta) is less than
w. Otherwise you're intersecting with the other edge of the rectangle.
But I'm guessing you have good reasons to believe that this will never
happen.
Roger.
On Fri, 16 Oct 2020 at 11:29, amundsen roald.baudoux@brutele.be wrote:
Hello,
I have a rectangle abcd with known width w and height h.
The rectangle is rotated according to the theta angle and becomes
ab'c'd'.
I can calculate the coordinates of the rotated rectangle. However my
knowledge in trigonometry is too weak to find the position of the
point p
on
the y axis. This point is at the intersection of the y-axis and the
line
passing through the upper side of the rectangle after rotation.
Does anyone know how to calculate this? The distance between a and b'
can
be
easily calculated, but from there I am not quite sure how to proceed.
http://forum.openscad.org/file/t2715/trigonometry_question.jpg
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** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110
**
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
OpenSCAD mailing list
Discuss@lists.openscad.org
http://lists.openscad.org/mailman/listinfo/discuss_lists.openscad.org
OpenSCAD mailing list
Discuss@lists.openscad.org
http://lists.openscad.org/mailman/listinfo/discuss_lists.openscad.org
--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110 **
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
Rogier,
By ab I mean the length of the line ab, which is just the short side of
the rectangle. The long side doesn't affect the position of p. You can
ignore the rectangle completely and just trig on triangle pab' once you
realise angle pab' is the same as theta.
On Fri, 16 Oct 2020 at 13:56, Rogier Wolff R.E.Wolff@bitwizard.nl wrote:
On Fri, Oct 16, 2020 at 12:04:26PM +0100, nop head wrote:
The triangle pab' is a right angle triangle and the angle pab' is equal
to
theta. So ab' / p = cos(theta). Hence p = ab' / cos(theta) and ab' = ab.
So
ab / cos(theta).
Although I might get a different formula... If I did things right the
result should be the same. It is possible that without pen-and-paper
I've done something wrong. And my trig is a bit rusty.
If you figure out a formula and want to double check it you can put in
the numbers that you can verify.
If my rectangle is 2 wide, 1 high and theta is 45deg. then the answer
should be sqrt (2).
if a is the width and b is the height, then ab/cos(theta) becomes 1.41
precisely the answer we expect....
But if our rectangle is 3 wide and the height and theta still 1 and 45
then the answer should still be 1.41 (sqrt(2)), but your formula gives
something else. The first example was a coincidence!
My formula however comes to yp = h * sqrt(1+1/tan(theta)^2) = sqrt(2)
for both examples. We can try a few more "known examples", but so far
my formula seems to do a bit better than yours.
Roger.
P.S. I started out thinking that both formulas would come out to the
same number but it seems that this is not the case.
On Fri, 16 Oct 2020 at 11:53, Rogier Wolff R.E.Wolff@bitwizard.nl
wrote:
On Fri, Oct 16, 2020 at 11:34:41AM +0100, nop head wrote:
I think it is just length(ab) / acos(theta)
Rotate the "whole figure" back along theta, but take the Y-axis along.
So now you want to find the intersection of the rotated Y-axis
and the upper side of your rectangle.
The line is: y/x = tan (theta).
and the topside of the rectangle is y = h
So I think you get h = tan (theta) * x or x = h/tan (theta)
So x,h is the intersection point, and now you calculate the lenght
from
the origin to find the Y-coordinate requested in the original question.
yp = sqrt (hh + h/tan (theta) h/tan (theta))
You could/should check that x = h/tan (theta) is less than
w. Otherwise you're intersecting with the other edge of the rectangle.
But I'm guessing you have good reasons to believe that this will never
happen.
Roger.
On Fri, 16 Oct 2020 at 11:29, amundsen roald.baudoux@brutele.be
wrote:
Hello,
I have a rectangle abcd with known width w and height h.
The rectangle is rotated according to the theta angle and becomes
ab'c'd'.
I can calculate the coordinates of the rotated rectangle. However
my
knowledge in trigonometry is too weak to find the position of the
point p
on
the y axis. This point is at the intersection of the y-axis and the
line
passing through the upper side of the rectangle after rotation.
Does anyone know how to calculate this? The distance between a and
b'
can
be
easily calculated, but from there I am not quite sure how to
proceed.
http://forum.openscad.org/file/t2715/trigonometry_question.jpg
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Discuss@lists.openscad.org
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+31-15-2049110
**
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233
**
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space
shuttle.
OpenSCAD mailing list
Discuss@lists.openscad.org
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--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110
**
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
OpenSCAD mailing list
Discuss@lists.openscad.org
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On Fri, Oct 16, 2020 at 03:33:45PM +0100, nop head wrote:
Rogier,
By ab I mean the length of the line ab, which is just the short side of
the rectangle. The long side doesn't affect the position of p. You can
ignore the rectangle completely and just trig on triangle pab' once you
realise angle pab' is the same as theta.
Hmm. Yeah looking at it that way I see your formula pop up
immediately... Seems I was doing something wrong. That's why I try to
check things with by putting in some "known" numbers...
Roger.
--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110 **
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.