Given visible landmarks of known location, and no magnetic compass, can you calculate your position?
$begingroup$
I have a bounty for this answer here
https://worldbuilding.stackexchange.com/questions/131610/what-would-be-used-for-coordinates-on-a-large-asteroid
The scenario: you are on an asteroid. Compasses do not work. You have 2 or more distant landmarks you can see. You have a map with all of these landmarks and the distances between each.
From your position, you cannot calculate a bearing relative to North (no compass) but you have a compass calipers and can calculate the angle to one landmark relative to another.
When I draw this I get one angle (your position and two distant landmarks) and the opposing side (known distance between the 2 landmarks). I don't think I can calculate the length of the other sides of the triangle. But if there are 3 more more landmarks, it looks to me like any given position within the triangle (or any other polygon made by the distant landmarks) is characterized by a series of triangles, each with an angle made up by my position and 2 of the distant landmarks.
Can I calculate my position with just the relative angles to distant landmarks of known position? No right triangles. No true north.
trigonometry
asked Dec 5 '18 at 14:21
WillkWillk
1085
$endgroup$
add a comment |
$begingroup$
I have a bounty for this answer here
https://worldbuilding.stackexchange.com/questions/131610/what-would-be-used-for-coordinates-on-a-large-asteroid
The scenario: you are on an asteroid. Compasses do not work. You have 2 or more distant landmarks you can see. You have a map with all of these landmarks and the distances between each.
From your position, you cannot calculate a bearing relative to North (no compass) but you have a compass calipers and can calculate the angle to one landmark relative to another.
When I draw this I get one angle (your position and two distant landmarks) and the opposing side (known distance between the 2 landmarks). I don't think I can calculate the length of the other sides of the triangle. But if there are 3 more more landmarks, it looks to me like any given position within the triangle (or any other polygon made by the distant landmarks) is characterized by a series of triangles, each with an angle made up by my position and 2 of the distant landmarks.
Can I calculate my position with just the relative angles to distant landmarks of known position? No right triangles. No true north.
trigonometry
asked Dec 5 '18 at 14:21
WillkWillk
1085
$endgroup$
2
$begingroup$
I note that zero and one are examples of "arbitrarily many". No such procedure can work for these choices of number of visible landmarks.
$endgroup$
– Eric Towers
Dec 5 '18 at 14:31
1
$begingroup$
I am surprised that "arbitrarily many" could be 0 but it could be 1. I am sure that neither 1 nor 0 landmarks will work. I have edited the question.
$endgroup$
– Willk
Dec 5 '18 at 14:44
$begingroup$
I’m not sure I totally understand the problem, but if your asteroid were a Euclidean plane (something like the spherical cow), then knowing the observed angular separation of two landmarks and the distance between them just places you on a certain circle passing through the two landmarks. With three landmarks, your location seems specified, save in exceptional cases. (You see that I’m assuming you can measure the angle subtended at your eye between two landmarks.)
$endgroup$
– Lubin
Dec 5 '18 at 15:06
$begingroup$
that's a couple of millennia old discipline : Triangulation
$endgroup$
– G Cab
Dec 5 '18 at 15:08
$begingroup$
I agree with Lubin. I believe in three-dimensional space knowing the angular separation places you on a sphere instead of a circle - so that still reduces down to a circle if you constrain to, say, a spherical asteroid. (The exceptional cases are probably a bit of a mess here and I think in practice you'd need four landmarks.)
$endgroup$
– Michael Lugo
Dec 5 '18 at 16:26
add a comment |
$begingroup$
I have a bounty for this answer here
https://worldbuilding.stackexchange.com/questions/131610/what-would-be-used-for-coordinates-on-a-large-asteroid
The scenario: you are on an asteroid. Compasses do not work. You have 2 or more distant landmarks you can see. You have a map with all of these landmarks and the distances between each.
From your position, you cannot calculate a bearing relative to North (no compass) but you have a compass calipers and can calculate the angle to one landmark relative to another.
When I draw this I get one angle (your position and two distant landmarks) and the opposing side (known distance between the 2 landmarks). I don't think I can calculate the length of the other sides of the triangle. But if there are 3 more more landmarks, it looks to me like any given position within the triangle (or any other polygon made by the distant landmarks) is characterized by a series of triangles, each with an angle made up by my position and 2 of the distant landmarks.
Can I calculate my position with just the relative angles to distant landmarks of known position? No right triangles. No true north.
trigonometry
asked Dec 5 '18 at 14:21
WillkWillk
1085
$endgroup$
I have a bounty for this answer here
https://worldbuilding.stackexchange.com/questions/131610/what-would-be-used-for-coordinates-on-a-large-asteroid
The scenario: you are on an asteroid. Compasses do not work. You have 2 or more distant landmarks you can see. You have a map with all of these landmarks and the distances between each.
From your position, you cannot calculate a bearing relative to North (no compass) but you have a compass calipers and can calculate the angle to one landmark relative to another.
When I draw this I get one angle (your position and two distant landmarks) and the opposing side (known distance between the 2 landmarks). I don't think I can calculate the length of the other sides of the triangle. But if there are 3 more more landmarks, it looks to me like any given position within the triangle (or any other polygon made by the distant landmarks) is characterized by a series of triangles, each with an angle made up by my position and 2 of the distant landmarks.
Can I calculate my position with just the relative angles to distant landmarks of known position? No right triangles. No true north.
trigonometry
trigonometry
asked Dec 5 '18 at 14:21
WillkWillk
1085
asked Dec 5 '18 at 14:21
WillkWillk
1085
edited Dec 5 '18 at 14:57
Willk
asked Dec 5 '18 at 14:21
WillkWillk
1085
asked Dec 5 '18 at 14:21
WillkWillk
1085
asked Dec 5 '18 at 14:21
WillkWillk
1085
1085
2
$begingroup$
I note that zero and one are examples of "arbitrarily many". No such procedure can work for these choices of number of visible landmarks.
$endgroup$
– Eric Towers
Dec 5 '18 at 14:31
1
$begingroup$
I am surprised that "arbitrarily many" could be 0 but it could be 1. I am sure that neither 1 nor 0 landmarks will work. I have edited the question.
$endgroup$
– Willk
Dec 5 '18 at 14:44
$begingroup$
I’m not sure I totally understand the problem, but if your asteroid were a Euclidean plane (something like the spherical cow), then knowing the observed angular separation of two landmarks and the distance between them just places you on a certain circle passing through the two landmarks. With three landmarks, your location seems specified, save in exceptional cases. (You see that I’m assuming you can measure the angle subtended at your eye between two landmarks.)
$endgroup$
– Lubin
Dec 5 '18 at 15:06
$begingroup$
that's a couple of millennia old discipline : Triangulation
$endgroup$
– G Cab
Dec 5 '18 at 15:08
$begingroup$
I agree with Lubin. I believe in three-dimensional space knowing the angular separation places you on a sphere instead of a circle - so that still reduces down to a circle if you constrain to, say, a spherical asteroid. (The exceptional cases are probably a bit of a mess here and I think in practice you'd need four landmarks.)
$endgroup$
– Michael Lugo
Dec 5 '18 at 16:26
add a comment |
2
$begingroup$
I note that zero and one are examples of "arbitrarily many". No such procedure can work for these choices of number of visible landmarks.
$endgroup$
– Eric Towers
Dec 5 '18 at 14:31
1
$begingroup$
I am surprised that "arbitrarily many" could be 0 but it could be 1. I am sure that neither 1 nor 0 landmarks will work. I have edited the question.
$endgroup$
– Willk
Dec 5 '18 at 14:44
$begingroup$
I’m not sure I totally understand the problem, but if your asteroid were a Euclidean plane (something like the spherical cow), then knowing the observed angular separation of two landmarks and the distance between them just places you on a certain circle passing through the two landmarks. With three landmarks, your location seems specified, save in exceptional cases. (You see that I’m assuming you can measure the angle subtended at your eye between two landmarks.)
$endgroup$
– Lubin
Dec 5 '18 at 15:06
$begingroup$
that's a couple of millennia old discipline : Triangulation
$endgroup$
– G Cab
Dec 5 '18 at 15:08
$begingroup$
I agree with Lubin. I believe in three-dimensional space knowing the angular separation places you on a sphere instead of a circle - so that still reduces down to a circle if you constrain to, say, a spherical asteroid. (The exceptional cases are probably a bit of a mess here and I think in practice you'd need four landmarks.)
$endgroup$
– Michael Lugo
Dec 5 '18 at 16:26
2
2
$begingroup$
I note that zero and one are examples of "arbitrarily many". No such procedure can work for these choices of number of visible landmarks.
$endgroup$
– Eric Towers
Dec 5 '18 at 14:31
$begingroup$
I note that zero and one are examples of "arbitrarily many". No such procedure can work for these choices of number of visible landmarks.
$endgroup$
– Eric Towers
Dec 5 '18 at 14:31
1
1
$begingroup$
I am surprised that "arbitrarily many" could be 0 but it could be 1. I am sure that neither 1 nor 0 landmarks will work. I have edited the question.
$endgroup$
– Willk
Dec 5 '18 at 14:44
$begingroup$
I am surprised that "arbitrarily many" could be 0 but it could be 1. I am sure that neither 1 nor 0 landmarks will work. I have edited the question.
$endgroup$
– Willk
Dec 5 '18 at 14:44
$begingroup$
I’m not sure I totally understand the problem, but if your asteroid were a Euclidean plane (something like the spherical cow), then knowing the observed angular separation of two landmarks and the distance between them just places you on a certain circle passing through the two landmarks. With three landmarks, your location seems specified, save in exceptional cases. (You see that I’m assuming you can measure the angle subtended at your eye between two landmarks.)
$endgroup$
– Lubin
Dec 5 '18 at 15:06
$begingroup$
I’m not sure I totally understand the problem, but if your asteroid were a Euclidean plane (something like the spherical cow), then knowing the observed angular separation of two landmarks and the distance between them just places you on a certain circle passing through the two landmarks. With three landmarks, your location seems specified, save in exceptional cases. (You see that I’m assuming you can measure the angle subtended at your eye between two landmarks.)
$endgroup$
– Lubin
Dec 5 '18 at 15:06
$begingroup$
that's a couple of millennia old discipline : Triangulation
$endgroup$
– G Cab
Dec 5 '18 at 15:08
$begingroup$
that's a couple of millennia old discipline : Triangulation
$endgroup$
– G Cab
Dec 5 '18 at 15:08
$begingroup$
I agree with Lubin. I believe in three-dimensional space knowing the angular separation places you on a sphere instead of a circle - so that still reduces down to a circle if you constrain to, say, a spherical asteroid. (The exceptional cases are probably a bit of a mess here and I think in practice you'd need four landmarks.)
$endgroup$
– Michael Lugo
Dec 5 '18 at 16:26
$begingroup$
I agree with Lubin. I believe in three-dimensional space knowing the angular separation places you on a sphere instead of a circle - so that still reduces down to a circle if you constrain to, say, a spherical asteroid. (The exceptional cases are probably a bit of a mess here and I think in practice you'd need four landmarks.)
$endgroup$
– Michael Lugo
Dec 5 '18 at 16:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.
We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$
Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.
Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
Spectacular! This is what I wanted; succinct and fine image too. Thank you very much. Should you be interested in rep on Worldbuilding, post this answer there and claim the 500.
$endgroup$
– Willk
Dec 5 '18 at 16:51
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%2f3027120%2fgiven-visible-landmarks-of-known-location-and-no-magnetic-compass-can-you-calc%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
$begingroup$
Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.
We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$
Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.
Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
Spectacular! This is what I wanted; succinct and fine image too. Thank you very much. Should you be interested in rep on Worldbuilding, post this answer there and claim the 500.
$endgroup$
– Willk
Dec 5 '18 at 16:51
add a comment |
$begingroup$
Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.
We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$
Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.
Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
Spectacular! This is what I wanted; succinct and fine image too. Thank you very much. Should you be interested in rep on Worldbuilding, post this answer there and claim the 500.
$endgroup$
– Willk
Dec 5 '18 at 16:51
add a comment |
$begingroup$
Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.
We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$
Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.
Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
$endgroup$
Yes, you can. Cover the asteroid with a triangular grid of points and measure their relative locations, making the coordinate system you want to locate yourself within. You need it fine enough that you can see the three nearest points wherever you are. Trivially, if you make it fine enough, just knowing which triangle you are in is accurate enough, but we assume the triangles are not that small.
We know points $A,B,C$ and distances $a,b,c$ from the map. We want to locate $D$ by measuring the angles $ADB, BDC, CDA$
Using the measured angle $ADB$ we know from the inscribed angle theorem that $D$ is on a circle centered at $E$ on the bisector of $AB$. If angle $ADB$ is greater than $frac pi 2$ the center of the circle is outside the triangle and located so the angle subtended by $AB$ is $2pi-2ADB$. If angle $ADB$ is less than $frac pi 2$ the center is inside the triangle and the angle subtended by $AB$ is $2ADB$. We can construct two of these circles based on different sides of triangle $ABC$, find the intersection, and that is point $D$. In the diagram, $E$ is the center of the circle that $D$ is on. We assume $ADB gt frac pi 2$. That will be true for at least two sides of the triangle.
Given the coordinates of $A,B$, we find $F$ as the midpoint of $AB$. Then $AEF=pi-ADB$ and $EF=frac 12ABtan (pi-ADB)$. $D$ is on the circle with center $E$ and radius $AE$. Do the same for another side, find the intersection, and you are done.
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
answered Dec 5 '18 at 16:37
Ross MillikanRoss Millikan
294k23198371
294k23198371
$begingroup$
Spectacular! This is what I wanted; succinct and fine image too. Thank you very much. Should you be interested in rep on Worldbuilding, post this answer there and claim the 500.
$endgroup$
– Willk
Dec 5 '18 at 16:51
add a comment |
$begingroup$
Spectacular! This is what I wanted; succinct and fine image too. Thank you very much. Should you be interested in rep on Worldbuilding, post this answer there and claim the 500.
$endgroup$
– Willk
Dec 5 '18 at 16:51
$begingroup$
Spectacular! This is what I wanted; succinct and fine image too. Thank you very much. Should you be interested in rep on Worldbuilding, post this answer there and claim the 500.
$endgroup$
– Willk
Dec 5 '18 at 16:51
$begingroup$
Spectacular! This is what I wanted; succinct and fine image too. Thank you very much. Should you be interested in rep on Worldbuilding, post this answer there and claim the 500.
$endgroup$
– Willk
Dec 5 '18 at 16:51
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%2f3027120%2fgiven-visible-landmarks-of-known-location-and-no-magnetic-compass-can-you-calc%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
2
$begingroup$
I note that zero and one are examples of "arbitrarily many". No such procedure can work for these choices of number of visible landmarks.
$endgroup$
– Eric Towers
Dec 5 '18 at 14:31
1
$begingroup$
I am surprised that "arbitrarily many" could be 0 but it could be 1. I am sure that neither 1 nor 0 landmarks will work. I have edited the question.
$endgroup$
– Willk
Dec 5 '18 at 14:44
$begingroup$
I’m not sure I totally understand the problem, but if your asteroid were a Euclidean plane (something like the spherical cow), then knowing the observed angular separation of two landmarks and the distance between them just places you on a certain circle passing through the two landmarks. With three landmarks, your location seems specified, save in exceptional cases. (You see that I’m assuming you can measure the angle subtended at your eye between two landmarks.)
$endgroup$
– Lubin
Dec 5 '18 at 15:06
$begingroup$
that's a couple of millennia old discipline : Triangulation
$endgroup$
– G Cab
Dec 5 '18 at 15:08
$begingroup$
I agree with Lubin. I believe in three-dimensional space knowing the angular separation places you on a sphere instead of a circle - so that still reduces down to a circle if you constrain to, say, a spherical asteroid. (The exceptional cases are probably a bit of a mess here and I think in practice you'd need four landmarks.)
$endgroup$
– Michael Lugo
Dec 5 '18 at 16:26