Mixing
Estimating the best geocoordinates for two different estimates
$begingroup$
I am having two sensors that give me two (latitude, longitude) geocoordinates. For both of them I also know their horizontal accuracy (radius in meters around the geocoordinates).
Let's express those estimates as (lat1, lon1, acc1) and (lat2, lon2, acc2).
I believe the statistical distribution around the two estimates is a gaussian function.
How can I compute the best output geocoordinates (lat0, lon0) from those two sources of information ? Moreover, how would be expressed the resulting accuracy acc0 ?
I was thinking using barycentric coordinates with weights (1/acc1) and (1/acc2) but I don't know if it's the best approach and it does not seem to work properly with lat, lon coordinates.
probability geometry
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
$endgroup$
add a comment |
$begingroup$
I am having two sensors that give me two (latitude, longitude) geocoordinates. For both of them I also know their horizontal accuracy (radius in meters around the geocoordinates).
Let's express those estimates as (lat1, lon1, acc1) and (lat2, lon2, acc2).
I believe the statistical distribution around the two estimates is a gaussian function.
How can I compute the best output geocoordinates (lat0, lon0) from those two sources of information ? Moreover, how would be expressed the resulting accuracy acc0 ?
I was thinking using barycentric coordinates with weights (1/acc1) and (1/acc2) but I don't know if it's the best approach and it does not seem to work properly with lat, lon coordinates.
probability geometry
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
$endgroup$
$begingroup$
If you consider concentric confidence circles at 90% , 95% , etc centered on your "nominal positions $P_1$ and $P_2$", if $O$ is the center of the sphere, reasoning on weighted (barycentric) mean along the arc of great circle obtained by intersection of plane $OP_1P_2$ with the sphere looks a good approach. The variance of the resulting distribution should be the barycenter of individual variances (only independence is needed ; no need to assume at first a gaussian behavior, even if it sounds plausible)
$endgroup$
– Jean Marie
Jan 19 at 22:53
add a comment |
$begingroup$
I am having two sensors that give me two (latitude, longitude) geocoordinates. For both of them I also know their horizontal accuracy (radius in meters around the geocoordinates).
Let's express those estimates as (lat1, lon1, acc1) and (lat2, lon2, acc2).
I believe the statistical distribution around the two estimates is a gaussian function.
How can I compute the best output geocoordinates (lat0, lon0) from those two sources of information ? Moreover, how would be expressed the resulting accuracy acc0 ?
I was thinking using barycentric coordinates with weights (1/acc1) and (1/acc2) but I don't know if it's the best approach and it does not seem to work properly with lat, lon coordinates.
probability geometry
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
$endgroup$
I am having two sensors that give me two (latitude, longitude) geocoordinates. For both of them I also know their horizontal accuracy (radius in meters around the geocoordinates).
Let's express those estimates as (lat1, lon1, acc1) and (lat2, lon2, acc2).
I believe the statistical distribution around the two estimates is a gaussian function.
How can I compute the best output geocoordinates (lat0, lon0) from those two sources of information ? Moreover, how would be expressed the resulting accuracy acc0 ?
I was thinking using barycentric coordinates with weights (1/acc1) and (1/acc2) but I don't know if it's the best approach and it does not seem to work properly with lat, lon coordinates.
probability geometry
probability geometry
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
asked Jan 19 at 20:52
michael-martinezmichael-martinez
1013
1013
$begingroup$
If you consider concentric confidence circles at 90% , 95% , etc centered on your "nominal positions $P_1$ and $P_2$", if $O$ is the center of the sphere, reasoning on weighted (barycentric) mean along the arc of great circle obtained by intersection of plane $OP_1P_2$ with the sphere looks a good approach. The variance of the resulting distribution should be the barycenter of individual variances (only independence is needed ; no need to assume at first a gaussian behavior, even if it sounds plausible)
$endgroup$
– Jean Marie
Jan 19 at 22:53
add a comment |
$begingroup$
If you consider concentric confidence circles at 90% , 95% , etc centered on your "nominal positions $P_1$ and $P_2$", if $O$ is the center of the sphere, reasoning on weighted (barycentric) mean along the arc of great circle obtained by intersection of plane $OP_1P_2$ with the sphere looks a good approach. The variance of the resulting distribution should be the barycenter of individual variances (only independence is needed ; no need to assume at first a gaussian behavior, even if it sounds plausible)
$endgroup$
– Jean Marie
Jan 19 at 22:53
$begingroup$
If you consider concentric confidence circles at 90% , 95% , etc centered on your "nominal positions $P_1$ and $P_2$", if $O$ is the center of the sphere, reasoning on weighted (barycentric) mean along the arc of great circle obtained by intersection of plane $OP_1P_2$ with the sphere looks a good approach. The variance of the resulting distribution should be the barycenter of individual variances (only independence is needed ; no need to assume at first a gaussian behavior, even if it sounds plausible)
$endgroup$
– Jean Marie
Jan 19 at 22:53
$begingroup$
If you consider concentric confidence circles at 90% , 95% , etc centered on your "nominal positions $P_1$ and $P_2$", if $O$ is the center of the sphere, reasoning on weighted (barycentric) mean along the arc of great circle obtained by intersection of plane $OP_1P_2$ with the sphere looks a good approach. The variance of the resulting distribution should be the barycenter of individual variances (only independence is needed ; no need to assume at first a gaussian behavior, even if it sounds plausible)
$endgroup$
– Jean Marie
Jan 19 at 22:53
add a comment |
0
active
oldest
votes
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%2f3079812%2festimating-the-best-geocoordinates-for-two-different-estimates%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3079812%2festimating-the-best-geocoordinates-for-two-different-estimates%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
$begingroup$
If you consider concentric confidence circles at 90% , 95% , etc centered on your "nominal positions $P_1$ and $P_2$", if $O$ is the center of the sphere, reasoning on weighted (barycentric) mean along the arc of great circle obtained by intersection of plane $OP_1P_2$ with the sphere looks a good approach. The variance of the resulting distribution should be the barycenter of individual variances (only independence is needed ; no need to assume at first a gaussian behavior, even if it sounds plausible)
$endgroup$
– Jean Marie
Jan 19 at 22:53