How to convert intersection of two lines into an arc?determine if 2 line segments are intersectingVerify that two line segments do not cross, or projected intersection is not on either lineAngle between two line segmentsFind out intersection of lines with only edge pointsHow can I calculate the angle between two lines on a sphere?How to calculate the radius of a Arc Segment given only the Arc Length and the Height of the arc segment?Find intersection of two lines given subtended angleHow to calculate 3D arc between two linesGiven a circle of radius r, and two points ('X' and 'Z') on that circle, can some circumcircular arc “XYZ” be constructed of length r?Length of a pivoted rectangle
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How to convert intersection of two lines into an arc?
determine if 2 line segments are intersectingVerify that two line segments do not cross, or projected intersection is not on either lineAngle between two line segmentsFind out intersection of lines with only edge pointsHow can I calculate the angle between two lines on a sphere?How to calculate the radius of a Arc Segment given only the Arc Length and the Height of the arc segment?Find intersection of two lines given subtended angleHow to calculate 3D arc between two linesGiven a circle of radius r, and two points ('X' and 'Z') on that circle, can some circumcircular arc “XYZ” be constructed of length r?Length of a pivoted rectangle
$begingroup$
I am struggling with finding coordinates of a point for arc origin.
I am programming a tool which converts a corner (intersection of two lines) into an arc of needed radius.
SETUP PICTURE
I have intersecting line segments AD and AC.
Known:
- Coordinates of Point A
- Length of Segment BE (r)
- Angle between AC and AB (which also equals to angle between AD & AB)
My problem is finding any dimension related to segment AB (length, projections, etc), which if known, I can solve the problem.
Ultimately Need:
- Points D, B, and C to construct an Arc and resize the intersecting lines to end at respective points C & D.
enter image description here
geometry
edited Mar 21 at 20:03
KReiser
10k21435
asked Mar 21 at 18:12
MicardMicard
31
$endgroup$
|
show 2 more comments
$begingroup$
I am struggling with finding coordinates of a point for arc origin.
I am programming a tool which converts a corner (intersection of two lines) into an arc of needed radius.
SETUP PICTURE
I have intersecting line segments AD and AC.
Known:
- Coordinates of Point A
- Length of Segment BE (r)
- Angle between AC and AB (which also equals to angle between AD & AB)
My problem is finding any dimension related to segment AB (length, projections, etc), which if known, I can solve the problem.
Ultimately Need:
- Points D, B, and C to construct an Arc and resize the intersecting lines to end at respective points C & D.
enter image description here
geometry
edited Mar 21 at 20:03
KReiser
10k21435
asked Mar 21 at 18:12
MicardMicard
31
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– clathratus
Mar 21 at 18:21
$begingroup$
What do you mean? I am not able to solve it correctly, that's all. I've used AEx projection + rx length to get the new coordinate of rx, but as you can see that would not be correct since it doesn't account for AB segment. I've no idea what to do to get info about the AB segment.
$endgroup$
– Micard
Mar 21 at 18:31
$begingroup$
Is $AC$ tangent to the circle at point $C$? if yes, then $AB=r(frac1sin angleEAC-1)$
$endgroup$
– Vasya
Mar 21 at 18:37
$begingroup$
Are AC and AD perpendicular, as shown in the drawing? Or could they meet at some angle other than 90 degrees?
$endgroup$
– John Hughes
Mar 21 at 18:46
$begingroup$
@Vasya Thanks, I will try that!
$endgroup$
– Micard
Mar 21 at 18:48
|
show 2 more comments
$begingroup$
I am struggling with finding coordinates of a point for arc origin.
I am programming a tool which converts a corner (intersection of two lines) into an arc of needed radius.
SETUP PICTURE
I have intersecting line segments AD and AC.
Known:
- Coordinates of Point A
- Length of Segment BE (r)
- Angle between AC and AB (which also equals to angle between AD & AB)
My problem is finding any dimension related to segment AB (length, projections, etc), which if known, I can solve the problem.
Ultimately Need:
- Points D, B, and C to construct an Arc and resize the intersecting lines to end at respective points C & D.
enter image description here
geometry
edited Mar 21 at 20:03
KReiser
10k21435
asked Mar 21 at 18:12
MicardMicard
31
$endgroup$
I am struggling with finding coordinates of a point for arc origin.
I am programming a tool which converts a corner (intersection of two lines) into an arc of needed radius.
SETUP PICTURE
I have intersecting line segments AD and AC.
Known:
- Coordinates of Point A
- Length of Segment BE (r)
- Angle between AC and AB (which also equals to angle between AD & AB)
My problem is finding any dimension related to segment AB (length, projections, etc), which if known, I can solve the problem.
Ultimately Need:
- Points D, B, and C to construct an Arc and resize the intersecting lines to end at respective points C & D.
enter image description here
geometry
geometry
edited Mar 21 at 20:03
KReiser
10k21435
asked Mar 21 at 18:12
MicardMicard
31
edited Mar 21 at 20:03
KReiser
10k21435
asked Mar 21 at 18:12
MicardMicard
31
edited Mar 21 at 20:03
KReiser
10k21435
edited Mar 21 at 20:03
KReiser
10k21435
edited Mar 21 at 20:03
KReiser
10k21435
10k21435
asked Mar 21 at 18:12
MicardMicard
31
asked Mar 21 at 18:12
MicardMicard
31
asked Mar 21 at 18:12
MicardMicard
31
31
$begingroup$
What have you tried?
$endgroup$
– clathratus
Mar 21 at 18:21
$begingroup$
What do you mean? I am not able to solve it correctly, that's all. I've used AEx projection + rx length to get the new coordinate of rx, but as you can see that would not be correct since it doesn't account for AB segment. I've no idea what to do to get info about the AB segment.
$endgroup$
– Micard
Mar 21 at 18:31
$begingroup$
Is $AC$ tangent to the circle at point $C$? if yes, then $AB=r(frac1sin angleEAC-1)$
$endgroup$
– Vasya
Mar 21 at 18:37
$begingroup$
Are AC and AD perpendicular, as shown in the drawing? Or could they meet at some angle other than 90 degrees?
$endgroup$
– John Hughes
Mar 21 at 18:46
$begingroup$
@Vasya Thanks, I will try that!
$endgroup$
– Micard
Mar 21 at 18:48
|
show 2 more comments
$begingroup$
What have you tried?
$endgroup$
– clathratus
Mar 21 at 18:21
$begingroup$
What do you mean? I am not able to solve it correctly, that's all. I've used AEx projection + rx length to get the new coordinate of rx, but as you can see that would not be correct since it doesn't account for AB segment. I've no idea what to do to get info about the AB segment.
$endgroup$
– Micard
Mar 21 at 18:31
$begingroup$
Is $AC$ tangent to the circle at point $C$? if yes, then $AB=r(frac1sin angleEAC-1)$
$endgroup$
– Vasya
Mar 21 at 18:37
$begingroup$
Are AC and AD perpendicular, as shown in the drawing? Or could they meet at some angle other than 90 degrees?
$endgroup$
– John Hughes
Mar 21 at 18:46
$begingroup$
@Vasya Thanks, I will try that!
$endgroup$
– Micard
Mar 21 at 18:48
$begingroup$
What have you tried?
$endgroup$
– clathratus
Mar 21 at 18:21
$begingroup$
What have you tried?
$endgroup$
– clathratus
Mar 21 at 18:21
$begingroup$
What do you mean? I am not able to solve it correctly, that's all. I've used AEx projection + rx length to get the new coordinate of rx, but as you can see that would not be correct since it doesn't account for AB segment. I've no idea what to do to get info about the AB segment.
$endgroup$
– Micard
Mar 21 at 18:31
$begingroup$
What do you mean? I am not able to solve it correctly, that's all. I've used AEx projection + rx length to get the new coordinate of rx, but as you can see that would not be correct since it doesn't account for AB segment. I've no idea what to do to get info about the AB segment.
$endgroup$
– Micard
Mar 21 at 18:31
$begingroup$
Is $AC$ tangent to the circle at point $C$? if yes, then $AB=r(frac1sin angleEAC-1)$
$endgroup$
– Vasya
Mar 21 at 18:37
$begingroup$
Is $AC$ tangent to the circle at point $C$? if yes, then $AB=r(frac1sin angleEAC-1)$
$endgroup$
– Vasya
Mar 21 at 18:37
$begingroup$
Are AC and AD perpendicular, as shown in the drawing? Or could they meet at some angle other than 90 degrees?
$endgroup$
– John Hughes
Mar 21 at 18:46
$begingroup$
Are AC and AD perpendicular, as shown in the drawing? Or could they meet at some angle other than 90 degrees?
$endgroup$
– John Hughes
Mar 21 at 18:46
$begingroup$
@Vasya Thanks, I will try that!
$endgroup$
– Micard
Mar 21 at 18:48
$begingroup$
@Vasya Thanks, I will try that!
$endgroup$
– Micard
Mar 21 at 18:48
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$triangleACE$ is a right triangle, therefore $$rover AE = sinangleCAE = sinleft(frac12angleCADright).$$ This lets you find $E$, and from there $B$, $C$ and $D$ are easily found. Note,too, that $angleAEC=fracpi2-angleCAE$.
answered Mar 21 at 19:14
amdamd
31.5k21052
$endgroup$
$begingroup$
Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem...
$endgroup$
– Micard
Mar 22 at 13:40
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
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oldest
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active
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votes
$begingroup$
$triangleACE$ is a right triangle, therefore $$rover AE = sinangleCAE = sinleft(frac12angleCADright).$$ This lets you find $E$, and from there $B$, $C$ and $D$ are easily found. Note,too, that $angleAEC=fracpi2-angleCAE$.
answered Mar 21 at 19:14
amdamd
31.5k21052
$endgroup$
$begingroup$
Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem...
$endgroup$
– Micard
Mar 22 at 13:40
add a comment |
$begingroup$
$triangleACE$ is a right triangle, therefore $$rover AE = sinangleCAE = sinleft(frac12angleCADright).$$ This lets you find $E$, and from there $B$, $C$ and $D$ are easily found. Note,too, that $angleAEC=fracpi2-angleCAE$.
answered Mar 21 at 19:14
amdamd
31.5k21052
$endgroup$
$begingroup$
Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem...
$endgroup$
– Micard
Mar 22 at 13:40
add a comment |
$begingroup$
$triangleACE$ is a right triangle, therefore $$rover AE = sinangleCAE = sinleft(frac12angleCADright).$$ This lets you find $E$, and from there $B$, $C$ and $D$ are easily found. Note,too, that $angleAEC=fracpi2-angleCAE$.
answered Mar 21 at 19:14
amdamd
31.5k21052
$endgroup$
$triangleACE$ is a right triangle, therefore $$rover AE = sinangleCAE = sinleft(frac12angleCADright).$$ This lets you find $E$, and from there $B$, $C$ and $D$ are easily found. Note,too, that $angleAEC=fracpi2-angleCAE$.
answered Mar 21 at 19:14
amdamd
31.5k21052
answered Mar 21 at 19:14
amdamd
31.5k21052
answered Mar 21 at 19:14
amdamd
31.5k21052
answered Mar 21 at 19:14
amdamd
31.5k21052
31.5k21052
$begingroup$
Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem...
$endgroup$
– Micard
Mar 22 at 13:40
add a comment |
$begingroup$
Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem...
$endgroup$
– Micard
Mar 22 at 13:40
$begingroup$
Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem...
$endgroup$
– Micard
Mar 22 at 13:40
$begingroup$
Thank you, works pretty well, not sure why I couldn't solve a 7th grade geometry problem...
$endgroup$
– Micard
Mar 22 at 13:40
add a comment |
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$begingroup$
What have you tried?
$endgroup$
– clathratus
Mar 21 at 18:21
$begingroup$
What do you mean? I am not able to solve it correctly, that's all. I've used AEx projection + rx length to get the new coordinate of rx, but as you can see that would not be correct since it doesn't account for AB segment. I've no idea what to do to get info about the AB segment.
$endgroup$
– Micard
Mar 21 at 18:31
$begingroup$
Is $AC$ tangent to the circle at point $C$? if yes, then $AB=r(frac1sin angleEAC-1)$
$endgroup$
– Vasya
Mar 21 at 18:37
$begingroup$
Are AC and AD perpendicular, as shown in the drawing? Or could they meet at some angle other than 90 degrees?
$endgroup$
– John Hughes
Mar 21 at 18:46
$begingroup$
@Vasya Thanks, I will try that!
$endgroup$
– Micard
Mar 21 at 18:48