Finding altitude and azimuth with an accelerometer and magnetometer
up vote
0
down vote
favorite
I posted this in the astronomy stack exchange forum, but considering that it is a very math intensive question I figured there could also be people on here that could help.
For a project with my astronomy professor, we are hooking up an array of 4 smaller telescopes to observe objects in space together.
To do this, we wanted to put an accelerometer and magnetometer on each telescope and have it read data from the gravitational pull on the accelerometer and the pull from the north magnetic pole on the magnetometer.
The accelerometer would measure the altitude, or the absolute height of measure point where accelerometer is at on the telescope. This would be converted to the angle of the telescope from its initial plane (when it makes a 90 degree angle with its stand and Cartesian z is zero).
The magnetometer would measure the pull of the of the magnetic north pole, and be able to give the azimuth angle, or the angle (from 0-360) that the telescope is pointing away from the magnetic north pole (0 degrees in the x direction would be pointing at the magnetic north pole).
Both readings are recorded in microteslas.
My professor said he was able to solve the altitude equation, but doesn't have time for the azimuth. He suggested treating the telescope itself as a 3D vector, normalizing it, and the projecting it down onto a flat plane, i.e. where it makes a 90 degree angle with the stand.
Would I use a rotational matrix for this? And should I treat the magnetometer and accelerometer vectors as two separate matrix vectors and use one to solve the other? What would the end equation look like?
linear-algebra trigonometry vector-spaces mathematical-astronomy
asked May 26 '16 at 19:38
Chris
11
add a comment |
up vote
0
down vote
favorite
I posted this in the astronomy stack exchange forum, but considering that it is a very math intensive question I figured there could also be people on here that could help.
For a project with my astronomy professor, we are hooking up an array of 4 smaller telescopes to observe objects in space together.
To do this, we wanted to put an accelerometer and magnetometer on each telescope and have it read data from the gravitational pull on the accelerometer and the pull from the north magnetic pole on the magnetometer.
The accelerometer would measure the altitude, or the absolute height of measure point where accelerometer is at on the telescope. This would be converted to the angle of the telescope from its initial plane (when it makes a 90 degree angle with its stand and Cartesian z is zero).
The magnetometer would measure the pull of the of the magnetic north pole, and be able to give the azimuth angle, or the angle (from 0-360) that the telescope is pointing away from the magnetic north pole (0 degrees in the x direction would be pointing at the magnetic north pole).
Both readings are recorded in microteslas.
My professor said he was able to solve the altitude equation, but doesn't have time for the azimuth. He suggested treating the telescope itself as a 3D vector, normalizing it, and the projecting it down onto a flat plane, i.e. where it makes a 90 degree angle with the stand.
Would I use a rotational matrix for this? And should I treat the magnetometer and accelerometer vectors as two separate matrix vectors and use one to solve the other? What would the end equation look like?
linear-algebra trigonometry vector-spaces mathematical-astronomy
asked May 26 '16 at 19:38
Chris
11
Do accelerometers measure gravitational attraction ? I don't know anything about it but I had assumed they measure acceleration.
– Simon
May 26 '16 at 20:38
Gravitational attraction is measured in acceleration. Earth's at sea level is roughly 9.8m/s/s, which is an acceleration. So yeah
– Chris
May 26 '16 at 21:08
Thanks - that's very interesting. I had the wrong idea. I had imagined they would only detect movement.
– Simon
May 26 '16 at 21:09
Accelerometers only measure proper acceleration which does not include gravity. When you think of an accelerometer, imagine two rocks connected by a spring, [x]~~[x], with some way to measure the deflection of the spring connecting them. In free fall, the rocks move in perfect tandem, so the spring deflection is zero. Place one rock on a table (with the other above) and the spring compresses. That compression will be due to the acceleration caused by the normal force from the table.
– jnez71
Mar 20 at 2:14
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I posted this in the astronomy stack exchange forum, but considering that it is a very math intensive question I figured there could also be people on here that could help.
For a project with my astronomy professor, we are hooking up an array of 4 smaller telescopes to observe objects in space together.
To do this, we wanted to put an accelerometer and magnetometer on each telescope and have it read data from the gravitational pull on the accelerometer and the pull from the north magnetic pole on the magnetometer.
The accelerometer would measure the altitude, or the absolute height of measure point where accelerometer is at on the telescope. This would be converted to the angle of the telescope from its initial plane (when it makes a 90 degree angle with its stand and Cartesian z is zero).
The magnetometer would measure the pull of the of the magnetic north pole, and be able to give the azimuth angle, or the angle (from 0-360) that the telescope is pointing away from the magnetic north pole (0 degrees in the x direction would be pointing at the magnetic north pole).
Both readings are recorded in microteslas.
My professor said he was able to solve the altitude equation, but doesn't have time for the azimuth. He suggested treating the telescope itself as a 3D vector, normalizing it, and the projecting it down onto a flat plane, i.e. where it makes a 90 degree angle with the stand.
Would I use a rotational matrix for this? And should I treat the magnetometer and accelerometer vectors as two separate matrix vectors and use one to solve the other? What would the end equation look like?
linear-algebra trigonometry vector-spaces mathematical-astronomy
asked May 26 '16 at 19:38
Chris
11
I posted this in the astronomy stack exchange forum, but considering that it is a very math intensive question I figured there could also be people on here that could help.
For a project with my astronomy professor, we are hooking up an array of 4 smaller telescopes to observe objects in space together.
To do this, we wanted to put an accelerometer and magnetometer on each telescope and have it read data from the gravitational pull on the accelerometer and the pull from the north magnetic pole on the magnetometer.
The accelerometer would measure the altitude, or the absolute height of measure point where accelerometer is at on the telescope. This would be converted to the angle of the telescope from its initial plane (when it makes a 90 degree angle with its stand and Cartesian z is zero).
The magnetometer would measure the pull of the of the magnetic north pole, and be able to give the azimuth angle, or the angle (from 0-360) that the telescope is pointing away from the magnetic north pole (0 degrees in the x direction would be pointing at the magnetic north pole).
Both readings are recorded in microteslas.
My professor said he was able to solve the altitude equation, but doesn't have time for the azimuth. He suggested treating the telescope itself as a 3D vector, normalizing it, and the projecting it down onto a flat plane, i.e. where it makes a 90 degree angle with the stand.
Would I use a rotational matrix for this? And should I treat the magnetometer and accelerometer vectors as two separate matrix vectors and use one to solve the other? What would the end equation look like?
linear-algebra trigonometry vector-spaces mathematical-astronomy
linear-algebra trigonometry vector-spaces mathematical-astronomy
asked May 26 '16 at 19:38
Chris
11
asked May 26 '16 at 19:38
Chris
11
asked May 26 '16 at 19:38
Chris
11
asked May 26 '16 at 19:38
Chris
11
asked May 26 '16 at 19:38
Chris
11
11
Do accelerometers measure gravitational attraction ? I don't know anything about it but I had assumed they measure acceleration.
– Simon
May 26 '16 at 20:38
Gravitational attraction is measured in acceleration. Earth's at sea level is roughly 9.8m/s/s, which is an acceleration. So yeah
– Chris
May 26 '16 at 21:08
Thanks - that's very interesting. I had the wrong idea. I had imagined they would only detect movement.
– Simon
May 26 '16 at 21:09
Accelerometers only measure proper acceleration which does not include gravity. When you think of an accelerometer, imagine two rocks connected by a spring, [x]~~[x], with some way to measure the deflection of the spring connecting them. In free fall, the rocks move in perfect tandem, so the spring deflection is zero. Place one rock on a table (with the other above) and the spring compresses. That compression will be due to the acceleration caused by the normal force from the table.
– jnez71
Mar 20 at 2:14
add a comment |
Do accelerometers measure gravitational attraction ? I don't know anything about it but I had assumed they measure acceleration.
– Simon
May 26 '16 at 20:38
Gravitational attraction is measured in acceleration. Earth's at sea level is roughly 9.8m/s/s, which is an acceleration. So yeah
– Chris
May 26 '16 at 21:08
Thanks - that's very interesting. I had the wrong idea. I had imagined they would only detect movement.
– Simon
May 26 '16 at 21:09
Accelerometers only measure proper acceleration which does not include gravity. When you think of an accelerometer, imagine two rocks connected by a spring, [x]~~[x], with some way to measure the deflection of the spring connecting them. In free fall, the rocks move in perfect tandem, so the spring deflection is zero. Place one rock on a table (with the other above) and the spring compresses. That compression will be due to the acceleration caused by the normal force from the table.
– jnez71
Mar 20 at 2:14
Do accelerometers measure gravitational attraction ? I don't know anything about it but I had assumed they measure acceleration.
– Simon
May 26 '16 at 20:38
Do accelerometers measure gravitational attraction ? I don't know anything about it but I had assumed they measure acceleration.
– Simon
May 26 '16 at 20:38
Gravitational attraction is measured in acceleration. Earth's at sea level is roughly 9.8m/s/s, which is an acceleration. So yeah
– Chris
May 26 '16 at 21:08
Gravitational attraction is measured in acceleration. Earth's at sea level is roughly 9.8m/s/s, which is an acceleration. So yeah
– Chris
May 26 '16 at 21:08
Thanks - that's very interesting. I had the wrong idea. I had imagined they would only detect movement.
– Simon
May 26 '16 at 21:09
Thanks - that's very interesting. I had the wrong idea. I had imagined they would only detect movement.
– Simon
May 26 '16 at 21:09
Accelerometers only measure proper acceleration which does not include gravity. When you think of an accelerometer, imagine two rocks connected by a spring, [x]~~[x], with some way to measure the deflection of the spring connecting them. In free fall, the rocks move in perfect tandem, so the spring deflection is zero. Place one rock on a table (with the other above) and the spring compresses. That compression will be due to the acceleration caused by the normal force from the table.
– jnez71
Mar 20 at 2:14
Accelerometers only measure proper acceleration which does not include gravity. When you think of an accelerometer, imagine two rocks connected by a spring, [x]~~[x], with some way to measure the deflection of the spring connecting them. In free fall, the rocks move in perfect tandem, so the spring deflection is zero. Place one rock on a table (with the other above) and the spring compresses. That compression will be due to the acceleration caused by the normal force from the table.
– jnez71
Mar 20 at 2:14
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
You can use a Rotation Matrix or alternatively by Vector Maths.
I'm assuming that the Accel and the Magnetometer are aligned in the same orientation, i.e. the X, Y and Z axis are pointing in the same directions, also, that the Z Axis is the Axis that you want to use for the Azimuth angle.
If you take your Vectors of G (Accel), H (Magnetometer) and Z (Axis of Azimuth Calc) values: $$ G = begin{bmatrix} Gx & Gy & Gz end{bmatrix} \ H = begin{bmatrix} Hx & Hy & Hz end{bmatrix} \ Z = begin{bmatrix} 0 & 0 & 1 end{bmatrix}
$$
You then need to Calculate:
$$ B = G times Z \
A = G times H $$
The Azimuth Angle $ theta $ is then:
$$ cos theta = frac{(A bullet B)}{lvert A rvert bullet lvert B rvert} $$
answered Nov 3 '16 at 11:13
Mike
186314
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',
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%2f1801253%2ffinding-altitude-and-azimuth-with-an-accelerometer-and-magnetometer%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
up vote
0
down vote
You can use a Rotation Matrix or alternatively by Vector Maths.
I'm assuming that the Accel and the Magnetometer are aligned in the same orientation, i.e. the X, Y and Z axis are pointing in the same directions, also, that the Z Axis is the Axis that you want to use for the Azimuth angle.
If you take your Vectors of G (Accel), H (Magnetometer) and Z (Axis of Azimuth Calc) values: $$ G = begin{bmatrix} Gx & Gy & Gz end{bmatrix} \ H = begin{bmatrix} Hx & Hy & Hz end{bmatrix} \ Z = begin{bmatrix} 0 & 0 & 1 end{bmatrix}
$$
You then need to Calculate:
$$ B = G times Z \
A = G times H $$
The Azimuth Angle $ theta $ is then:
$$ cos theta = frac{(A bullet B)}{lvert A rvert bullet lvert B rvert} $$
answered Nov 3 '16 at 11:13
Mike
186314
add a comment |
up vote
0
down vote
You can use a Rotation Matrix or alternatively by Vector Maths.
I'm assuming that the Accel and the Magnetometer are aligned in the same orientation, i.e. the X, Y and Z axis are pointing in the same directions, also, that the Z Axis is the Axis that you want to use for the Azimuth angle.
If you take your Vectors of G (Accel), H (Magnetometer) and Z (Axis of Azimuth Calc) values: $$ G = begin{bmatrix} Gx & Gy & Gz end{bmatrix} \ H = begin{bmatrix} Hx & Hy & Hz end{bmatrix} \ Z = begin{bmatrix} 0 & 0 & 1 end{bmatrix}
$$
You then need to Calculate:
$$ B = G times Z \
A = G times H $$
The Azimuth Angle $ theta $ is then:
$$ cos theta = frac{(A bullet B)}{lvert A rvert bullet lvert B rvert} $$
answered Nov 3 '16 at 11:13
Mike
186314
add a comment |
up vote
0
down vote
up vote
0
down vote
You can use a Rotation Matrix or alternatively by Vector Maths.
I'm assuming that the Accel and the Magnetometer are aligned in the same orientation, i.e. the X, Y and Z axis are pointing in the same directions, also, that the Z Axis is the Axis that you want to use for the Azimuth angle.
If you take your Vectors of G (Accel), H (Magnetometer) and Z (Axis of Azimuth Calc) values: $$ G = begin{bmatrix} Gx & Gy & Gz end{bmatrix} \ H = begin{bmatrix} Hx & Hy & Hz end{bmatrix} \ Z = begin{bmatrix} 0 & 0 & 1 end{bmatrix}
$$
You then need to Calculate:
$$ B = G times Z \
A = G times H $$
The Azimuth Angle $ theta $ is then:
$$ cos theta = frac{(A bullet B)}{lvert A rvert bullet lvert B rvert} $$
answered Nov 3 '16 at 11:13
Mike
186314
You can use a Rotation Matrix or alternatively by Vector Maths.
I'm assuming that the Accel and the Magnetometer are aligned in the same orientation, i.e. the X, Y and Z axis are pointing in the same directions, also, that the Z Axis is the Axis that you want to use for the Azimuth angle.
If you take your Vectors of G (Accel), H (Magnetometer) and Z (Axis of Azimuth Calc) values: $$ G = begin{bmatrix} Gx & Gy & Gz end{bmatrix} \ H = begin{bmatrix} Hx & Hy & Hz end{bmatrix} \ Z = begin{bmatrix} 0 & 0 & 1 end{bmatrix}
$$
You then need to Calculate:
$$ B = G times Z \
A = G times H $$
The Azimuth Angle $ theta $ is then:
$$ cos theta = frac{(A bullet B)}{lvert A rvert bullet lvert B rvert} $$
answered Nov 3 '16 at 11:13
Mike
186314
answered Nov 3 '16 at 11:13
Mike
186314
answered Nov 3 '16 at 11:13
Mike
186314
answered Nov 3 '16 at 11:13
Mike
186314
186314
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f1801253%2ffinding-altitude-and-azimuth-with-an-accelerometer-and-magnetometer%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
Do accelerometers measure gravitational attraction ? I don't know anything about it but I had assumed they measure acceleration.
– Simon
May 26 '16 at 20:38
Gravitational attraction is measured in acceleration. Earth's at sea level is roughly 9.8m/s/s, which is an acceleration. So yeah
– Chris
May 26 '16 at 21:08
Thanks - that's very interesting. I had the wrong idea. I had imagined they would only detect movement.
– Simon
May 26 '16 at 21:09
Accelerometers only measure proper acceleration which does not include gravity. When you think of an accelerometer, imagine two rocks connected by a spring, [x]~~[x], with some way to measure the deflection of the spring connecting them. In free fall, the rocks move in perfect tandem, so the spring deflection is zero. Place one rock on a table (with the other above) and the spring compresses. That compression will be due to the acceleration caused by the normal force from the table.
– jnez71
Mar 20 at 2:14