Calculate the true diameter of stars from photographic plateIs the light we see from stars extremely...
Why is 'diphthong' pronounced the way it is?
How do you voice extended chords?
Best way to strengthen a wheel?
Why is a temp table a more efficient solution to the Halloween Problem than an eager spool?
How can I play a serial killer in a party of good PCs?
Why zero tolerance on nudity in space?
How can 若い時 and ずっと work in this sentence?
Count repetitions of an array
Possible issue with my W4 and tax return
Sprint is 2 week and 40-stories
The No-Straight Maze
Stuck on a Geometry Puzzle
Does diversity provide anything that meritocracy does not?
How do you funnel food off a cutting board?
Can the "Friends" spell be used without making the target hostile?
Why didn't Tom Riddle take the presence of Fawkes and the Sorting Hat as more of a threat?
What is the difference between "...", '...', $'...', and $"..." quotes?
How does Leonard in "Memento" remember reading and writing?
Equivalent of "illegal" for violating civil law
Does the ditching switch allow an A320 to float indefinitely?
False written accusations not made public - is there law to cover this?
How to not let the Identify spell spoil everything?
"Starve to death" Vs. "Starve to the point of death"
What can I do to encourage my players to use their consumables?
Calculate the true diameter of stars from photographic plate
Is the light we see from stars extremely old?Calculate Distance To StarsHow to calculate Longitude from Right Ascension?Diameter of any galaxyAre there stars in the first photo from space?How to calculate position of an unknown star knowing positions of some other stars from an image?Is the diameter of the observable universe a relative quantity?Diameter of Astronomical Object Using MagnificationCalculating the Diameter of Jupiter through ImageUnderstanding The Turnover Point of Angular Diameter Distance
$begingroup$
First of all, sorry if this is the wrong place to ask this, and in this case, please guide me to the correct place. (And sorry for the tags, I really have no idea which ones to choose)
I am reading the book "Fourier Analysis" from T. W. Körner, and in chapter 95, he explains how you can compute the diameter of stars on a photographic plate. He says that "Since observations of the nearest stars at six-monthly intervals (i.e. using a diameter of the earth's orbit as a surveyor's base line) enable astronomers to measure the distance of these stars, knowledge of the apparent diameter (i.e. the diameters of the discs
on the photographic plate) would then enable us to calculate the true diameters of the nearest stars".
So how do you calculate the true diameter of a star given its apparent diameter on a photographic plate? Is it related to the "zooming capacity" of the telescope one uses?
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
observational-astronomy distances photography
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
First of all, sorry if this is the wrong place to ask this, and in this case, please guide me to the correct place. (And sorry for the tags, I really have no idea which ones to choose)
I am reading the book "Fourier Analysis" from T. W. Körner, and in chapter 95, he explains how you can compute the diameter of stars on a photographic plate. He says that "Since observations of the nearest stars at six-monthly intervals (i.e. using a diameter of the earth's orbit as a surveyor's base line) enable astronomers to measure the distance of these stars, knowledge of the apparent diameter (i.e. the diameters of the discs
on the photographic plate) would then enable us to calculate the true diameters of the nearest stars".
So how do you calculate the true diameter of a star given its apparent diameter on a photographic plate? Is it related to the "zooming capacity" of the telescope one uses?
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
observational-astronomy distances photography
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
First of all, sorry if this is the wrong place to ask this, and in this case, please guide me to the correct place. (And sorry for the tags, I really have no idea which ones to choose)
I am reading the book "Fourier Analysis" from T. W. Körner, and in chapter 95, he explains how you can compute the diameter of stars on a photographic plate. He says that "Since observations of the nearest stars at six-monthly intervals (i.e. using a diameter of the earth's orbit as a surveyor's base line) enable astronomers to measure the distance of these stars, knowledge of the apparent diameter (i.e. the diameters of the discs
on the photographic plate) would then enable us to calculate the true diameters of the nearest stars".
So how do you calculate the true diameter of a star given its apparent diameter on a photographic plate? Is it related to the "zooming capacity" of the telescope one uses?
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
observational-astronomy distances photography
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
First of all, sorry if this is the wrong place to ask this, and in this case, please guide me to the correct place. (And sorry for the tags, I really have no idea which ones to choose)
I am reading the book "Fourier Analysis" from T. W. Körner, and in chapter 95, he explains how you can compute the diameter of stars on a photographic plate. He says that "Since observations of the nearest stars at six-monthly intervals (i.e. using a diameter of the earth's orbit as a surveyor's base line) enable astronomers to measure the distance of these stars, knowledge of the apparent diameter (i.e. the diameters of the discs
on the photographic plate) would then enable us to calculate the true diameters of the nearest stars".
So how do you calculate the true diameter of a star given its apparent diameter on a photographic plate? Is it related to the "zooming capacity" of the telescope one uses?
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
observational-astronomy distances photography
observational-astronomy distances photography
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Laurent HayezLaurent Hayez
1184
1184
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Laurent Hayez is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.
In principle, this is basic trigonometry: you have a measured angle ($alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear diameter of the star ($d$, in the same units as the distance) it's a simple matter of $sin alpha = d/D$. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just $alpha approx d/D rightarrow d = alpha D$ (assuming $alpha$ is in radians).
As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by stellar parallax, which involves the same basic trigonometric argument (only this time you know the "diameter", which is the size of the Earth's orbit).
So why is this, as I said, not generally possible? In practice, the diameter of the telescope sets a lower limit on observed angular sizes -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter $approx$ that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of $sim 0.5$ arc seconds (full-width half-maximum) or worse at the best sites. Since the largest stars in angular terms have diameters of about 0.05 arc seconds, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the same observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.
It is possible to do better for bright, nearby stars using large telescopes in space or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the "stars on photographic plates" scenario your book seems to envisage.
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an angular diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.
$endgroup$
$begingroup$
Ahhh the angular diameter makes much more sense! Thank you for the explanation. So for instance, in this paper apps.dtic.mil/dtic/tr/fulltext/u2/a067077.pdf, they measure the diameter of $alpha$ Boo (I think this is Arcturus), and using your formula, I obain a "true" diameter of $sim 4.37 times 10^{10}$ vs $sim 3.53 times 10^{10}$ from Wikipedia. Does it look correct to you? Thank you again for the great answer!
$endgroup$
– Laurent Hayez
54 mins ago
$begingroup$
P.S.: The book I'm reading explains the technique introduced by Labeyrie in 1970 ( adsabs.harvard.edu/full/1970A%26A.....6...85L ) to go around the problem that the blurring due to atmospheric effects is greater that the apparent diameter. I think he obtains the linear diameter that way, and from the plate scale he can deduce angular diameter. The book states that more than 30 stars had their diameters measured in this way :-).
$endgroup$
– Laurent Hayez
32 mins ago
$begingroup$
@LaurentHayez When you say something has a value of such-and-such you need to state the units. Without units it has no meaning, it's just a number.
$endgroup$
– StephenG
29 mins ago
$begingroup$
@StephenG You're right. A bad habit I have from mathematics :-). I used meters in the first comment.
$endgroup$
– Laurent Hayez
24 mins ago
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: "514"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
});
}
});
Laurent Hayez is a new contributor. Be nice, and check out our Code of Conduct.
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%2fastronomy.stackexchange.com%2fquestions%2f29763%2fcalculate-the-true-diameter-of-stars-from-photographic-plate%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
$begingroup$
Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.
In principle, this is basic trigonometry: you have a measured angle ($alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear diameter of the star ($d$, in the same units as the distance) it's a simple matter of $sin alpha = d/D$. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just $alpha approx d/D rightarrow d = alpha D$ (assuming $alpha$ is in radians).
As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by stellar parallax, which involves the same basic trigonometric argument (only this time you know the "diameter", which is the size of the Earth's orbit).
So why is this, as I said, not generally possible? In practice, the diameter of the telescope sets a lower limit on observed angular sizes -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter $approx$ that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of $sim 0.5$ arc seconds (full-width half-maximum) or worse at the best sites. Since the largest stars in angular terms have diameters of about 0.05 arc seconds, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the same observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.
It is possible to do better for bright, nearby stars using large telescopes in space or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the "stars on photographic plates" scenario your book seems to envisage.
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an angular diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.
$endgroup$
$begingroup$
Ahhh the angular diameter makes much more sense! Thank you for the explanation. So for instance, in this paper apps.dtic.mil/dtic/tr/fulltext/u2/a067077.pdf, they measure the diameter of $alpha$ Boo (I think this is Arcturus), and using your formula, I obain a "true" diameter of $sim 4.37 times 10^{10}$ vs $sim 3.53 times 10^{10}$ from Wikipedia. Does it look correct to you? Thank you again for the great answer!
$endgroup$
– Laurent Hayez
54 mins ago
$begingroup$
P.S.: The book I'm reading explains the technique introduced by Labeyrie in 1970 ( adsabs.harvard.edu/full/1970A%26A.....6...85L ) to go around the problem that the blurring due to atmospheric effects is greater that the apparent diameter. I think he obtains the linear diameter that way, and from the plate scale he can deduce angular diameter. The book states that more than 30 stars had their diameters measured in this way :-).
$endgroup$
– Laurent Hayez
32 mins ago
$begingroup$
@LaurentHayez When you say something has a value of such-and-such you need to state the units. Without units it has no meaning, it's just a number.
$endgroup$
– StephenG
29 mins ago
$begingroup$
@StephenG You're right. A bad habit I have from mathematics :-). I used meters in the first comment.
$endgroup$
– Laurent Hayez
24 mins ago
add a comment |
$begingroup$
Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.
In principle, this is basic trigonometry: you have a measured angle ($alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear diameter of the star ($d$, in the same units as the distance) it's a simple matter of $sin alpha = d/D$. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just $alpha approx d/D rightarrow d = alpha D$ (assuming $alpha$ is in radians).
As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by stellar parallax, which involves the same basic trigonometric argument (only this time you know the "diameter", which is the size of the Earth's orbit).
So why is this, as I said, not generally possible? In practice, the diameter of the telescope sets a lower limit on observed angular sizes -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter $approx$ that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of $sim 0.5$ arc seconds (full-width half-maximum) or worse at the best sites. Since the largest stars in angular terms have diameters of about 0.05 arc seconds, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the same observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.
It is possible to do better for bright, nearby stars using large telescopes in space or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the "stars on photographic plates" scenario your book seems to envisage.
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an angular diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.
$endgroup$
$begingroup$
Ahhh the angular diameter makes much more sense! Thank you for the explanation. So for instance, in this paper apps.dtic.mil/dtic/tr/fulltext/u2/a067077.pdf, they measure the diameter of $alpha$ Boo (I think this is Arcturus), and using your formula, I obain a "true" diameter of $sim 4.37 times 10^{10}$ vs $sim 3.53 times 10^{10}$ from Wikipedia. Does it look correct to you? Thank you again for the great answer!
$endgroup$
– Laurent Hayez
54 mins ago
$begingroup$
P.S.: The book I'm reading explains the technique introduced by Labeyrie in 1970 ( adsabs.harvard.edu/full/1970A%26A.....6...85L ) to go around the problem that the blurring due to atmospheric effects is greater that the apparent diameter. I think he obtains the linear diameter that way, and from the plate scale he can deduce angular diameter. The book states that more than 30 stars had their diameters measured in this way :-).
$endgroup$
– Laurent Hayez
32 mins ago
$begingroup$
@LaurentHayez When you say something has a value of such-and-such you need to state the units. Without units it has no meaning, it's just a number.
$endgroup$
– StephenG
29 mins ago
$begingroup$
@StephenG You're right. A bad habit I have from mathematics :-). I used meters in the first comment.
$endgroup$
– Laurent Hayez
24 mins ago
add a comment |
$begingroup$
Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.
In principle, this is basic trigonometry: you have a measured angle ($alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear diameter of the star ($d$, in the same units as the distance) it's a simple matter of $sin alpha = d/D$. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just $alpha approx d/D rightarrow d = alpha D$ (assuming $alpha$ is in radians).
As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by stellar parallax, which involves the same basic trigonometric argument (only this time you know the "diameter", which is the size of the Earth's orbit).
So why is this, as I said, not generally possible? In practice, the diameter of the telescope sets a lower limit on observed angular sizes -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter $approx$ that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of $sim 0.5$ arc seconds (full-width half-maximum) or worse at the best sites. Since the largest stars in angular terms have diameters of about 0.05 arc seconds, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the same observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.
It is possible to do better for bright, nearby stars using large telescopes in space or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the "stars on photographic plates" scenario your book seems to envisage.
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an angular diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.
$endgroup$
Let's pretend for the moment that this is possible (it's generally not, as I'll explain below) and see how we would go about it.
In principle, this is basic trigonometry: you have a measured angle ($alpha$, the diameter of the star), you know the distance to the star ($D$, from the six-month-separated parallax measurements), so to determine the linear diameter of the star ($d$, in the same units as the distance) it's a simple matter of $sin alpha = d/D$. In practice, the angles are small enough that you can use the small-angle approximation, so it's really just $alpha approx d/D rightarrow d = alpha D$ (assuming $alpha$ is in radians).
As your book notes, you need to know the distance to the star; for nearby stars, this can be obtained by stellar parallax, which involves the same basic trigonometric argument (only this time you know the "diameter", which is the size of the Earth's orbit).
So why is this, as I said, not generally possible? In practice, the diameter of the telescope sets a lower limit on observed angular sizes -- anything with a true diameter smaller than the telescope's angular limit will appear to have an angular diameter $approx$ that limit. In addition, any telescope underneath the Earth's atmosphere will suffer the effects of blurring due to atmospheric turbulence, which sets an angular diameter limit of $sim 0.5$ arc seconds (full-width half-maximum) or worse at the best sites. Since the largest stars in angular terms have diameters of about 0.05 arc seconds, the result is that all stars in the image (photographic plate, CCD camera, etc.) will have the same observed angular diameter (e.g., 0.5 arc seconds at a really good site), regardless of their true diameters.
It is possible to do better for bright, nearby stars using large telescopes in space or special interferometric techniques -- see the second link in the previous paragraph for an example -- but this is going beyond the "stars on photographic plates" scenario your book seems to envisage.
And if for instance, using a 100'' telescope, you measure that a star has an apparent diameter of 0.66mm, can you directly compute the true diameter of the star?
Leaving aside the problems mentioned above, you don't have enough information. You need to convert the linear diameter in mm to an angular diameter, which means you need to know the plate scale (e.g., arcsec/mm) of the particular telescope + camera setup. You also need to know the distance to the star.
answered 2 hours ago
Peter ErwinPeter Erwin
4,501724
4,501724
$begingroup$
Ahhh the angular diameter makes much more sense! Thank you for the explanation. So for instance, in this paper apps.dtic.mil/dtic/tr/fulltext/u2/a067077.pdf, they measure the diameter of $alpha$ Boo (I think this is Arcturus), and using your formula, I obain a "true" diameter of $sim 4.37 times 10^{10}$ vs $sim 3.53 times 10^{10}$ from Wikipedia. Does it look correct to you? Thank you again for the great answer!
$endgroup$
– Laurent Hayez
54 mins ago
$begingroup$
P.S.: The book I'm reading explains the technique introduced by Labeyrie in 1970 ( adsabs.harvard.edu/full/1970A%26A.....6...85L ) to go around the problem that the blurring due to atmospheric effects is greater that the apparent diameter. I think he obtains the linear diameter that way, and from the plate scale he can deduce angular diameter. The book states that more than 30 stars had their diameters measured in this way :-).
$endgroup$
– Laurent Hayez
32 mins ago
$begingroup$
@LaurentHayez When you say something has a value of such-and-such you need to state the units. Without units it has no meaning, it's just a number.
$endgroup$
– StephenG
29 mins ago
$begingroup$
@StephenG You're right. A bad habit I have from mathematics :-). I used meters in the first comment.
$endgroup$
– Laurent Hayez
24 mins ago
add a comment |
$begingroup$
Ahhh the angular diameter makes much more sense! Thank you for the explanation. So for instance, in this paper apps.dtic.mil/dtic/tr/fulltext/u2/a067077.pdf, they measure the diameter of $alpha$ Boo (I think this is Arcturus), and using your formula, I obain a "true" diameter of $sim 4.37 times 10^{10}$ vs $sim 3.53 times 10^{10}$ from Wikipedia. Does it look correct to you? Thank you again for the great answer!
$endgroup$
– Laurent Hayez
54 mins ago
$begingroup$
P.S.: The book I'm reading explains the technique introduced by Labeyrie in 1970 ( adsabs.harvard.edu/full/1970A%26A.....6...85L ) to go around the problem that the blurring due to atmospheric effects is greater that the apparent diameter. I think he obtains the linear diameter that way, and from the plate scale he can deduce angular diameter. The book states that more than 30 stars had their diameters measured in this way :-).
$endgroup$
– Laurent Hayez
32 mins ago
$begingroup$
@LaurentHayez When you say something has a value of such-and-such you need to state the units. Without units it has no meaning, it's just a number.
$endgroup$
– StephenG
29 mins ago
$begingroup$
@StephenG You're right. A bad habit I have from mathematics :-). I used meters in the first comment.
$endgroup$
– Laurent Hayez
24 mins ago
$begingroup$
Ahhh the angular diameter makes much more sense! Thank you for the explanation. So for instance, in this paper apps.dtic.mil/dtic/tr/fulltext/u2/a067077.pdf, they measure the diameter of $alpha$ Boo (I think this is Arcturus), and using your formula, I obain a "true" diameter of $sim 4.37 times 10^{10}$ vs $sim 3.53 times 10^{10}$ from Wikipedia. Does it look correct to you? Thank you again for the great answer!
$endgroup$
– Laurent Hayez
54 mins ago
$begingroup$
Ahhh the angular diameter makes much more sense! Thank you for the explanation. So for instance, in this paper apps.dtic.mil/dtic/tr/fulltext/u2/a067077.pdf, they measure the diameter of $alpha$ Boo (I think this is Arcturus), and using your formula, I obain a "true" diameter of $sim 4.37 times 10^{10}$ vs $sim 3.53 times 10^{10}$ from Wikipedia. Does it look correct to you? Thank you again for the great answer!
$endgroup$
– Laurent Hayez
54 mins ago
$begingroup$
P.S.: The book I'm reading explains the technique introduced by Labeyrie in 1970 ( adsabs.harvard.edu/full/1970A%26A.....6...85L ) to go around the problem that the blurring due to atmospheric effects is greater that the apparent diameter. I think he obtains the linear diameter that way, and from the plate scale he can deduce angular diameter. The book states that more than 30 stars had their diameters measured in this way :-).
$endgroup$
– Laurent Hayez
32 mins ago
$begingroup$
P.S.: The book I'm reading explains the technique introduced by Labeyrie in 1970 ( adsabs.harvard.edu/full/1970A%26A.....6...85L ) to go around the problem that the blurring due to atmospheric effects is greater that the apparent diameter. I think he obtains the linear diameter that way, and from the plate scale he can deduce angular diameter. The book states that more than 30 stars had their diameters measured in this way :-).
$endgroup$
– Laurent Hayez
32 mins ago
$begingroup$
@LaurentHayez When you say something has a value of such-and-such you need to state the units. Without units it has no meaning, it's just a number.
$endgroup$
– StephenG
29 mins ago
$begingroup$
@LaurentHayez When you say something has a value of such-and-such you need to state the units. Without units it has no meaning, it's just a number.
$endgroup$
– StephenG
29 mins ago
$begingroup$
@StephenG You're right. A bad habit I have from mathematics :-). I used meters in the first comment.
$endgroup$
– Laurent Hayez
24 mins ago
$begingroup$
@StephenG You're right. A bad habit I have from mathematics :-). I used meters in the first comment.
$endgroup$
– Laurent Hayez
24 mins ago
add a comment |
Laurent Hayez is a new contributor. Be nice, and check out our Code of Conduct.
Laurent Hayez is a new contributor. Be nice, and check out our Code of Conduct.
Laurent Hayez is a new contributor. Be nice, and check out our Code of Conduct.
Laurent Hayez is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Astronomy 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%2fastronomy.stackexchange.com%2fquestions%2f29763%2fcalculate-the-true-diameter-of-stars-from-photographic-plate%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