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Calculate the true diameter of stars from photographic plate


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3












$begingroup$


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?










share|improve this question







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.







$endgroup$

















    3












    $begingroup$


    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?










    share|improve this question







    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.







    $endgroup$















      3












      3








      3





      $begingroup$


      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?










      share|improve this question







      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






      share|improve this question







      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.











      share|improve this question







      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.









      share|improve this question




      share|improve this question






      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.






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          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.






          share|improve this answer









          $endgroup$













          • $begingroup$
            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











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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.






          share|improve this answer









          $endgroup$













          • $begingroup$
            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
















          4












          $begingroup$

          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.






          share|improve this answer









          $endgroup$













          • $begingroup$
            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














          4












          4








          4





          $begingroup$

          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.






          share|improve this answer









          $endgroup$



          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.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          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


















          • $begingroup$
            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










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