Is it possible to identify a circular arc given its length and two of its endpoints?
$begingroup$
Suppose you have the length of the circular arc AB, in addition to the coordinates of A and B. Is this information sufficient to draw the arc (i.e. find its center point)?
geometry circle arc-length
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
$endgroup$
add a comment |
$begingroup$
Suppose you have the length of the circular arc AB, in addition to the coordinates of A and B. Is this information sufficient to draw the arc (i.e. find its center point)?
geometry circle arc-length
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
$endgroup$
$begingroup$
Only if the arc length is equal to the distance between the two points. In that case the arc is the straight line segment connecting the two points.
$endgroup$
– Hans Engler
Dec 12 '18 at 13:52
$begingroup$
Is it given that the arc is of a circle?
$endgroup$
– Martund
Dec 12 '18 at 14:02
$begingroup$
Yes, the arc is a circular arc. I will edit accordingly.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:02
$begingroup$
Sorry, I also have the coordinates of A and B, not only the distance between them.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:06
add a comment |
$begingroup$
Suppose you have the length of the circular arc AB, in addition to the coordinates of A and B. Is this information sufficient to draw the arc (i.e. find its center point)?
geometry circle arc-length
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
$endgroup$
Suppose you have the length of the circular arc AB, in addition to the coordinates of A and B. Is this information sufficient to draw the arc (i.e. find its center point)?
geometry circle arc-length
geometry circle arc-length
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
edited Dec 12 '18 at 14:03
Wais Kamal
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
asked Dec 12 '18 at 13:46
Wais KamalWais Kamal
1356
1356
$begingroup$
Only if the arc length is equal to the distance between the two points. In that case the arc is the straight line segment connecting the two points.
$endgroup$
– Hans Engler
Dec 12 '18 at 13:52
$begingroup$
Is it given that the arc is of a circle?
$endgroup$
– Martund
Dec 12 '18 at 14:02
$begingroup$
Yes, the arc is a circular arc. I will edit accordingly.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:02
$begingroup$
Sorry, I also have the coordinates of A and B, not only the distance between them.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:06
add a comment |
$begingroup$
Only if the arc length is equal to the distance between the two points. In that case the arc is the straight line segment connecting the two points.
$endgroup$
– Hans Engler
Dec 12 '18 at 13:52
$begingroup$
Is it given that the arc is of a circle?
$endgroup$
– Martund
Dec 12 '18 at 14:02
$begingroup$
Yes, the arc is a circular arc. I will edit accordingly.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:02
$begingroup$
Sorry, I also have the coordinates of A and B, not only the distance between them.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:06
$begingroup$
Only if the arc length is equal to the distance between the two points. In that case the arc is the straight line segment connecting the two points.
$endgroup$
– Hans Engler
Dec 12 '18 at 13:52
$begingroup$
Only if the arc length is equal to the distance between the two points. In that case the arc is the straight line segment connecting the two points.
$endgroup$
– Hans Engler
Dec 12 '18 at 13:52
$begingroup$
Is it given that the arc is of a circle?
$endgroup$
– Martund
Dec 12 '18 at 14:02
$begingroup$
Is it given that the arc is of a circle?
$endgroup$
– Martund
Dec 12 '18 at 14:02
$begingroup$
Yes, the arc is a circular arc. I will edit accordingly.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:02
$begingroup$
Yes, the arc is a circular arc. I will edit accordingly.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:02
$begingroup$
Sorry, I also have the coordinates of A and B, not only the distance between them.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:06
$begingroup$
Sorry, I also have the coordinates of A and B, not only the distance between them.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is sufficient except for an ambiguity of which side of the line segment $AB$ the center lies on. The center lies on the bisector of $AB$. If $r$ is the radius of the circle, $L$ is the length of the arc, $theta$ is the angle at the center subtended by $AB$, $d$ is the length of $AB$ we have $$L=rtheta\d=2rsinleft (frac theta 2right)$$ and we can solve these numerically to get $r, theta$. This lets you find possible points for the center.
A figure is below. My $B,C$ are your $A,B$.
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
Thank you, may you explain where you got the formula from?
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:45
$begingroup$
Draw the figure with the center of the circle, the two radii to $A,B$, and the line from the center to the midpoint of $AB$. You form two right triangles, which is where the sine comes from. I added a figure
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:47
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036705%2fis-it-possible-to-identify-a-circular-arc-given-its-length-and-two-of-its-endpoi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is sufficient except for an ambiguity of which side of the line segment $AB$ the center lies on. The center lies on the bisector of $AB$. If $r$ is the radius of the circle, $L$ is the length of the arc, $theta$ is the angle at the center subtended by $AB$, $d$ is the length of $AB$ we have $$L=rtheta\d=2rsinleft (frac theta 2right)$$ and we can solve these numerically to get $r, theta$. This lets you find possible points for the center.
A figure is below. My $B,C$ are your $A,B$.
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
Thank you, may you explain where you got the formula from?
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:45
$begingroup$
Draw the figure with the center of the circle, the two radii to $A,B$, and the line from the center to the midpoint of $AB$. You form two right triangles, which is where the sine comes from. I added a figure
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:47
add a comment |
$begingroup$
It is sufficient except for an ambiguity of which side of the line segment $AB$ the center lies on. The center lies on the bisector of $AB$. If $r$ is the radius of the circle, $L$ is the length of the arc, $theta$ is the angle at the center subtended by $AB$, $d$ is the length of $AB$ we have $$L=rtheta\d=2rsinleft (frac theta 2right)$$ and we can solve these numerically to get $r, theta$. This lets you find possible points for the center.
A figure is below. My $B,C$ are your $A,B$.
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
Thank you, may you explain where you got the formula from?
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:45
$begingroup$
Draw the figure with the center of the circle, the two radii to $A,B$, and the line from the center to the midpoint of $AB$. You form two right triangles, which is where the sine comes from. I added a figure
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:47
add a comment |
$begingroup$
It is sufficient except for an ambiguity of which side of the line segment $AB$ the center lies on. The center lies on the bisector of $AB$. If $r$ is the radius of the circle, $L$ is the length of the arc, $theta$ is the angle at the center subtended by $AB$, $d$ is the length of $AB$ we have $$L=rtheta\d=2rsinleft (frac theta 2right)$$ and we can solve these numerically to get $r, theta$. This lets you find possible points for the center.
A figure is below. My $B,C$ are your $A,B$.
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
$endgroup$
It is sufficient except for an ambiguity of which side of the line segment $AB$ the center lies on. The center lies on the bisector of $AB$. If $r$ is the radius of the circle, $L$ is the length of the arc, $theta$ is the angle at the center subtended by $AB$, $d$ is the length of $AB$ we have $$L=rtheta\d=2rsinleft (frac theta 2right)$$ and we can solve these numerically to get $r, theta$. This lets you find possible points for the center.
A figure is below. My $B,C$ are your $A,B$.
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
edited Dec 12 '18 at 14:51
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
answered Dec 12 '18 at 14:43
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
Thank you, may you explain where you got the formula from?
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:45
$begingroup$
Draw the figure with the center of the circle, the two radii to $A,B$, and the line from the center to the midpoint of $AB$. You form two right triangles, which is where the sine comes from. I added a figure
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:47
add a comment |
$begingroup$
Thank you, may you explain where you got the formula from?
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:45
$begingroup$
Draw the figure with the center of the circle, the two radii to $A,B$, and the line from the center to the midpoint of $AB$. You form two right triangles, which is where the sine comes from. I added a figure
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:47
$begingroup$
Thank you, may you explain where you got the formula from?
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:45
$begingroup$
Thank you, may you explain where you got the formula from?
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:45
$begingroup$
Draw the figure with the center of the circle, the two radii to $A,B$, and the line from the center to the midpoint of $AB$. You form two right triangles, which is where the sine comes from. I added a figure
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:47
$begingroup$
Draw the figure with the center of the circle, the two radii to $A,B$, and the line from the center to the midpoint of $AB$. You form two right triangles, which is where the sine comes from. I added a figure
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:47
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036705%2fis-it-possible-to-identify-a-circular-arc-given-its-length-and-two-of-its-endpoi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Only if the arc length is equal to the distance between the two points. In that case the arc is the straight line segment connecting the two points.
$endgroup$
– Hans Engler
Dec 12 '18 at 13:52
$begingroup$
Is it given that the arc is of a circle?
$endgroup$
– Martund
Dec 12 '18 at 14:02
$begingroup$
Yes, the arc is a circular arc. I will edit accordingly.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:02
$begingroup$
Sorry, I also have the coordinates of A and B, not only the distance between them.
$endgroup$
– Wais Kamal
Dec 12 '18 at 14:06