Why is short-wave infrared portion of electromagnetic spectrum so sensitive to fire?












4















After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus







share|improve this question




















  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    Mar 22 at 8:34








  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    Mar 22 at 8:48
















4















After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus







share|improve this question




















  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    Mar 22 at 8:34








  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    Mar 22 at 8:48














4












4








4








After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus







share|improve this question
















After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus




remote-sensing copernicus






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 22 at 8:35









PolyGeo♦

53.8k1781244




53.8k1781244










asked Mar 22 at 7:57









Robert BuckleyRobert Buckley

4,8651255115




4,8651255115








  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    Mar 22 at 8:34








  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    Mar 22 at 8:48














  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    Mar 22 at 8:34








  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    Mar 22 at 8:48








1




1





This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

– PolyGeo♦
Mar 22 at 8:34







This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

– PolyGeo♦
Mar 22 at 8:34






2




2





@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

– radouxju
Mar 22 at 8:48





@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

– radouxju
Mar 22 at 8:48










2 Answers
2






active

oldest

votes


















9














According to Wien's law, the wavelength of the peak emission depends on the temperature. 



wavelength_of_peak = 2.898*10^(-3)/T


where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






share|improve this answer


























  • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

    – MaxW
    Mar 22 at 22:24



















4














First, I would reword the question to "Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?"



Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:





  • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


  • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.





So, the spectrum is not sensitive, the fact is that temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






share|improve this answer










New contributor




gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "79"
    };
    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
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fgis.stackexchange.com%2fquestions%2f316341%2fwhy-is-short-wave-infrared-portion-of-electromagnetic-spectrum-so-sensitive-to-f%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9














    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






    share|improve this answer


























    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      Mar 22 at 22:24
















    9














    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






    share|improve this answer


























    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      Mar 22 at 22:24














    9












    9








    9







    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






    share|improve this answer















    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 22 at 8:49

























    answered Mar 22 at 8:36









    radouxjuradouxju

    41.2k144120




    41.2k144120













    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      Mar 22 at 22:24



















    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      Mar 22 at 22:24

















    Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

    – MaxW
    Mar 22 at 22:24





    Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

    – MaxW
    Mar 22 at 22:24













    4














    First, I would reword the question to "Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?"



    Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:





    • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


    • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.





    So, the spectrum is not sensitive, the fact is that temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



    source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






    share|improve this answer










    New contributor




    gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.

























      4














      First, I would reword the question to "Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?"



      Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:





      • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


      • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.





      So, the spectrum is not sensitive, the fact is that temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



      source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






      share|improve this answer










      New contributor




      gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.























        4












        4








        4







        First, I would reword the question to "Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?"



        Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:





        • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


        • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.





        So, the spectrum is not sensitive, the fact is that temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



        source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






        share|improve this answer










        New contributor




        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.










        First, I would reword the question to "Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?"



        Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:





        • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


        • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.





        So, the spectrum is not sensitive, the fact is that temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



        source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law







        share|improve this answer










        New contributor




        gustavovelascoh 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 answer



        share|improve this answer








        edited 16 hours ago





















        New contributor




        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered Mar 22 at 14:54









        gustavovelascohgustavovelascoh

        1414




        1414




        New contributor




        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Geographic Information Systems 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.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fgis.stackexchange.com%2fquestions%2f316341%2fwhy-is-short-wave-infrared-portion-of-electromagnetic-spectrum-so-sensitive-to-f%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more