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Re: [GPSL] Earths shadow

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  • Joe
    8 MIN PER WHAT? Sig The Original Rolling Ball Clock Idle Tyme Idle-Tyme.com http://www.idle-tyme.com
    Message 1 of 13 , Feb 8, 2013
      8 MIN PER WHAT?

      Sig
      The Original Rolling Ball Clock
      Idle Tyme
      Idle-Tyme.com
      http://www.idle-tyme.com
      On 2/8/2013 6:46 PM, Mike Manes wrote:
      Hmmm ... assuming the sun's disc subtends 2 degrees and
      the earth rotates 15 deg per hour, then a SWAG suggests
      2/15 of an hour, or about 8 minutes.
      73 de Mike W5VSI
      
      On 2/8/13 5:39 PM, James Ewen wrote:
      
      On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...> wrote:
      
      
      Does anyone know how fst does the earths shadow rise after sundown?
      
      How about describe what it is that you are trying to describe. The
      earth's shadow doesn't rise. There is a day/night terminator that
      travels across the face of the earth. That is dependent upon the
      latitude that you are observing, and the time of year.
      
      Seeing as how you have just launched a balloon, you're probably
      interested in figuring out when the balloon will cross into daylight
      if it was launched after sunset, or when will the balloon stop being
      sunlit, and fall into the earth's shadow if the balloon is currently
      sunlit.
      
      That again will be a factor of latitude and time of year, and also the
      altitude that you are interested in. Probably the term you are looking
      for is local sunset at a specific altitude.
      
      Some of the smart guys here know the formula... I think Mark Conner
      was posting some information like that a while ago.
      
      --
      James
      VE6SRV
      
      
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    • Joe
      I was just hopeing it went back into shadow soon. it launched a bit earlier than planned and even tho it was like 30 min after sundown on the ground it became
      Message 2 of 13 , Feb 8, 2013
        I was just hopeing it went back into shadow soon. it launched a bit earlier than planned and even tho it was like 30 min after sundown on the ground it became illuminated after launch and I wanted it to get dark before it nears float altitude.

        On 2/8/2013 6:48 PM, L. Paul Verhage wrote:

        Earth's shadow rises above the local horizon. It fades out as it rises higher. You won't see much of it by 30 minutes after sunset. It's also visible about 30 minutes prior to sunrise.

        It's more visible from near space. Several of my videos show it.

        Paul

        On Feb 8, 2013 5:39 PM, "James Ewen" <ve6srv@...> wrote:
        On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...> wrote:

        > Does anyone know how fst does the earths shadow rise after sundown?

        How about describe what it is that you are trying to describe. The
        earth's shadow doesn't rise. There is a day/night terminator that
        travels across the face of the earth. That is dependent upon the
        latitude that you are observing, and the time of year.

        Seeing as how you have just launched a balloon, you're probably
        interested in figuring out when the balloon will cross into daylight
        if it was launched after sunset, or when will the balloon stop being
        sunlit, and fall into the earth's shadow if the balloon is currently
        sunlit.

        That again will be a factor of latitude and time of year, and also the
        altitude that you are interested in. Probably the term you are looking
        for is local sunset at a specific altitude.

        Some of the smart guys here know the formula... I think Mark Conner
        was posting some information like that a while ago.

        --
        James
        VE6SRV


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      • Mark Conner
        It wasn t me, and probably by the time I sort out all the math I m pretty sure the balloon will be in darkness. This would have been interesting to see. On
        Message 3 of 13 , Feb 8, 2013
          It wasn't me, and probably by the time I sort out all the math I'm pretty sure the balloon will be in darkness.

          This would have been interesting to see.  On the pre-sunrise flights I've participated in, the balloon becomes brilliant very quickly as you're going up into daylight as sunrise is "moving down" towards the surface, so you have a rapid transition from darkness to full illumination.  For a post-sunset flight, you're chasing the earth's shadow moving up in the atmosphere, overtaking it and going into the light, then as the angles get more shallow the shadow will move up more rapidly and overtake the balloon again.  The math on this could get pretty complex.

          73 de Mark N9XTN

          On Fri, Feb 8, 2013 at 6:39 PM, James Ewen <ve6srv@...> wrote:
          On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...> wrote:

          That again will be a factor of latitude and time of year, and also the
          altitude that you are interested in. Probably the term you are looking
          for is local sunset at a specific altitude.

          Some of the smart guys here know the formula... I think Mark Conner
          was posting some information like that a while ago.

          --
          James
          VE6SRV

        • Joe
          agreed! Sig The Original Rolling Ball Clock Idle Tyme Idle-Tyme.com http://www.idle-tyme.com
          Message 4 of 13 , Feb 8, 2013
            agreed!
            Sig
            The Original Rolling Ball Clock
            Idle Tyme
            Idle-Tyme.com
            http://www.idle-tyme.com
            On 2/8/2013 7:18 PM, Mark Conner wrote:
            It wasn't me, and probably by the time I sort out all the math I'm pretty sure the balloon will be in darkness.

            This would have been interesting to see.  On the pre-sunrise flights I've participated in, the balloon becomes brilliant very quickly as you're going up into daylight as sunrise is "moving down" towards the surface, so you have a rapid transition from darkness to full illumination.  For a post-sunset flight, you're chasing the earth's shadow moving up in the atmosphere, overtaking it and going into the light, then as the angles get more shallow the shadow will move up more rapidly and overtake the balloon again.  The math on this could get pretty complex.

            73 de Mark N9XTN

            On Fri, Feb 8, 2013 at 6:39 PM, James Ewen <ve6srv@...> wrote:
            On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...> wrote:

            That again will be a factor of latitude and time of year, and also the
            altitude that you are interested in. Probably the term you are looking
            for is local sunset at a specific altitude.

            Some of the smart guys here know the formula... I think Mark Conner
            was posting some information like that a while ago.

            --
            James
            VE6SRV


          • L. Paul Verhage
            Here s now I figured when to launch a balloon to see sunrise. Make adujstments to the numbers to suit your particular situation. Using the equation distance
            Message 5 of 13 , Feb 9, 2013
              Here's now I figured when to launch a balloon to see sunrise. Make adujstments to the numbers to suit your particular situation.

              Using the equation distance (miles) = square root [1.5 * altitude (feet)], the horizon is 387 miles at an altitude of 100,000 feet. Earth rotates once every 24 hours and Earth's circumference at a latitude of 40 degrees is 19,151 miles (25,000 miles * cos(40 deg)). So Earth rotates about 798 miles per hour at 40 deg N. If the near spacecraft sees the horizon 387 miles away, it will observe sunrise 29 minutes earlier than the ground below it.

              I figure the balloon will take 90 minutes to get to 100,000 feet and I want the balloon there 29 minutes before local sunrise. So the launch must occur 1 hour 59 minutes prior to local sunrise.    

              I have three flights that observed sunrise in near space. However, I think it would be more interesting to dawn in near space. So my next attempt will try to observe the horizon perhaps 30 minutes prior to sunrise.

              Paul

              On Fri, Feb 8, 2013 at 6:54 PM, Joe <nss@...> wrote:
              agreed!

              The Original Rolling Ball Clock
              Idle Tyme
              Idle-Tyme.com
              http://www.idle-tyme.com
              On 2/8/2013 7:18 PM, Mark Conner wrote:
              It wasn't me, and probably by the time I sort out all the math I'm pretty sure the balloon will be in darkness.

              This would have been interesting to see.  On the pre-sunrise flights I've participated in, the balloon becomes brilliant very quickly as you're going up into daylight as sunrise is "moving down" towards the surface, so you have a rapid transition from darkness to full illumination.  For a post-sunset flight, you're chasing the earth's shadow moving up in the atmosphere, overtaking it and going into the light, then as the angles get more shallow the shadow will move up more rapidly and overtake the balloon again.  The math on this could get pretty complex.

              73 de Mark N9XTN

              On Fri, Feb 8, 2013 at 6:39 PM, James Ewen <ve6srv@...> wrote:
              On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...> wrote:

              That again will be a factor of latitude and time of year, and also the
              altitude that you are interested in. Probably the term you are looking
              for is local sunset at a specific altitude.

              Some of the smart guys here know the formula... I think Mark Conner
              was posting some information like that a while ago.

              --
              James
              VE6SRV





              --
              Onwards and Upwards,
              Paul
            • Joe
              This is great. And matches Pretty close to what my Astronomy program shows. I used the center of the Sun, If at 3 meters elevation the sun this morning rises
              Message 6 of 13 , Feb 10, 2013
                This is great.
                And matches  Pretty close to what my Astronomy program shows.

                I used the center of the Sun, If at 3 meters elevation the sun this morning rises at 7:03 AM

                At 27,000 Meters the center of the sun is at the horizon at 6:45 IE: 18 minutes earlier

                and at 34,000 meters the sun is centered on the horizon at 6:40  IE: 23 minutes earlier.

                What brought this all up was we launched the balloon earlier that planned, On the ground the Sun had set. But shortly after launch the balloon suddenly lit up. It was fully illuminated by the Sun Yikes!  Didn't want that.
                And the thought was

                Will this eventually get back into the earths shadow? Or will the rate of rise out pace the rate the earths shadow rises. So that brought up the thought of how fast does the shadow actually rise?

                I am willing to bet it matters on day of year, and the locations latitude also.  But sadly my math skills are no where up for this calculation.

                I wonder if it is even linear. I'm imagining in my mind that it is not. I'm thinking is rises slower when it is low, and much faster as it gets higher.


                Anyone?

                Joe WB9SBD

                Sig
                The Original Rolling Ball Clock
                Idle Tyme
                Idle-Tyme.com
                http://www.idle-tyme.com
                On 2/9/2013 8:49 PM, L. Paul Verhage wrote:
                Here's now I figured when to launch a balloon to see sunrise. Make adujstments to the numbers to suit your particular situation.

                Using the equation distance (miles) = square root [1.5 * altitude (feet)], the horizon is 387 miles at an altitude of 100,000 feet. Earth rotates once every 24 hours and Earth's circumference at a latitude of 40 degrees is 19,151 miles (25,000 miles * cos(40 deg)). So Earth rotates about 798 miles per hour at 40 deg N. If the near spacecraft sees the horizon 387 miles away, it will observe sunrise 29 minutes earlier than the ground below it.

                I figure the balloon will take 90 minutes to get to 100,000 feet and I want the balloon there 29 minutes before local sunrise. So the launch must occur 1 hour 59 minutes prior to local sunrise.    

                I have three flights that observed sunrise in near space. However, I think it would be more interesting to dawn in near space. So my next attempt will try to observe the horizon perhaps 30 minutes prior to sunrise.

                Paul

                On Fri, Feb 8, 2013 at 6:54 PM, Joe <nss@...> wrote:
                agreed!

                The Original Rolling Ball Clock
                Idle Tyme
                Idle-Tyme.com
                http://www.idle-tyme.com
                On 2/8/2013 7:18 PM, Mark Conner wrote:
                It wasn't me, and probably by the time I sort out all the math I'm pretty sure the balloon will be in darkness.

                This would have been interesting to see.  On the pre-sunrise flights I've participated in, the balloon becomes brilliant very quickly as you're going up into daylight as sunrise is "moving down" towards the surface, so you have a rapid transition from darkness to full illumination.  For a post-sunset flight, you're chasing the earth's shadow moving up in the atmosphere, overtaking it and going into the light, then as the angles get more shallow the shadow will move up more rapidly and overtake the balloon again.  The math on this could get pretty complex.

                73 de Mark N9XTN

                On Fri, Feb 8, 2013 at 6:39 PM, James Ewen <ve6srv@...> wrote:
                On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...>wrote:

                That again will be a factor of latitude and time of year, and also the
                altitude that you are interested in. Probably the term you are looking
                for is local sunset at a specific altitude.

                Some of the smart guys here know the formula... I think Mark Conner
                was posting some information like that a while ago.

                --
                James
                VE6SRV





                --
                Onwards and Upwards,
                Paul

              • L. Paul Verhage
                The sun has to set faster. It takes the balloon roughly 90 minutes of climbing to see the sun set 20 minutes later. It can t be a linear function since both
                Message 7 of 13 , Feb 10, 2013

                  The sun has to set faster. It takes the balloon roughly 90 minutes of climbing to see the sun set 20 minutes later.

                  It can't be a linear function since both the horizon recedes and the sun sets later by the square root if the altitude. Since the altitude is proportion to time aloft, the delay in observing sunset is proportion is proportional to mission elapse time.

                  To make everything linear and hold the sun's altitude above the horizon constant, we must change the ascent rate from a constant to function with an increasing value. The balloon has to climb at a faster rate.

                  I think if the function describing the climb rate was proportional to the square root of the mission elapse time, the two multiplied square roots would cancel to give you a constant for the delay in seeing sunset. In that case, the sun maintains its altitude above the horizon.

                  I'll try to write up a mathematic explanation of my reasoning and post it.

                  Paul

                  On Feb 10, 2013 7:28 AM, "Joe" <nss@...> wrote:
                  This is great.
                  And matches  Pretty close to what my Astronomy program shows.

                  I used the center of the Sun, If at 3 meters elevation the sun this morning rises at 7:03 AM

                  At 27,000 Meters the center of the sun is at the horizon at 6:45 IE: 18 minutes earlier

                  and at 34,000 meters the sun is centered on the horizon at 6:40  IE: 23 minutes earlier.

                  What brought this all up was we launched the balloon earlier that planned, On the ground the Sun had set. But shortly after launch the balloon suddenly lit up. It was fully illuminated by the Sun Yikes!  Didn't want that.
                  And the thought was

                  Will this eventually get back into the earths shadow? Or will the rate of rise out pace the rate the earths shadow rises. So that brought up the thought of how fast does the shadow actually rise?

                  I am willing to bet it matters on day of year, and the locations latitude also.  But sadly my math skills are no where up for this calculation.

                  I wonder if it is even linear. I'm imagining in my mind that it is not. I'm thinking is rises slower when it is low, and much faster as it gets higher.


                  Anyone?

                  Joe WB9SBD


                  The Original Rolling Ball Clock
                  Idle Tyme
                  Idle-Tyme.com
                  http://www.idle-tyme.com
                  On 2/9/2013 8:49 PM, L. Paul Verhage wrote:


                  Here's now I figured when to launch a balloon to see sunrise. Make adujstments to the numbers to suit your particular situation.

                  Using the equation distance (miles) = square root [1.5 * altitude (feet)], the horizon is 387 miles at an altitude of 100,000 feet. Earth rotates once every 24 hours and Earth's circumference at a latitude of 40 degrees is 19,151 miles (25,000 miles * cos(40 deg)). So Earth rotates about 798 miles per hour at 40 deg N. If the near spacecraft sees the horizon 387 miles away, it will observe sunrise 29 minutes earlier than the ground below it.

                  I figure the balloon will take 90 minutes to get to 100,000 feet and I want the balloon there 29 minutes before local sunrise. So the launch must occur 1 hour 59 minutes prior to local sunrise.    

                  I have three flights that observed sunrise in near space. However, I think it would be more interesting to dawn in near space. So my next attempt will try to observe the horizon perhaps 30 minutes prior to sunrise.

                  Paul

                  On Fri, Feb 8, 2013 at 6:54 PM, Joe <nss@...> wrote:
                  agreed!

                  The Original Rolling Ball Clock
                  Idle Tyme
                  Idle-Tyme.com
                  http://www.idle-tyme.com
                  On 2/8/2013 7:18 PM, Mark Conner wrote:
                  It wasn't me, and probably by the time I sort out all the math I'm pretty sure the balloon will be in darkness.

                  This would have been interesting to see.  On the pre-sunrise flights I've participated in, the balloon becomes brilliant very quickly as you're going up into daylight as sunrise is "moving down" towards the surface, so you have a rapid transition from darkness to full illumination.  For a post-sunset flight, you're chasing the earth's shadow moving up in the atmosphere, overtaking it and going into the light, then as the angles get more shallow the shadow will move up more rapidly and overtake the balloon again.  The math on this could get pretty complex.

                  73 de Mark N9XTN

                  On Fri, Feb 8, 2013 at 6:39 PM, James Ewen <ve6srv@...> wrote:
                  On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...>wrote:

                  That again will be a factor of latitude and time of year, and also the
                  altitude that you are interested in. Probably the term you are looking
                  for is local sunset at a specific altitude.

                  Some of the smart guys here know the formula... I think Mark Conner
                  was posting some information like that a while ago.

                  --
                  James
                  VE6SRV





                  --
                  Onwards and Upwards,
                  Paul


                • Don Fraser
                  Hi Joe; The day/night line can be easily seen using a program called GeoClock. See http://www.mygeoclock.com/geoclock/ for a sample and you ll see a sample.
                  Message 8 of 13 , Feb 10, 2013
                    Sig

                    Hi Joe;

                     

                    The day/night line can be easily seen using a program called GeoClock.  See http://www.mygeoclock.com/geoclock/ for a sample and you’ll see a sample.  During the year the day/night line moves up in winter and down in summer reflecting the shorter or longer days. 

                     

                    Back in the old days (ok late 70s) there were mechanical versions of the display in the NORAD command post.

                     

                    Don Fraser, WA9WWS

                    Home: 719-598-6174

                    Cell:  715-896-4319

                     

                    From: GPSL@yahoogroups.com [mailto:GPSL@yahoogroups.com] On Behalf Of Joe
                    Sent: Sunday, February 10, 2013 7:28 AM
                    To: L. Paul Verhage
                    Cc: GPSL@yahoogroups.com
                    Subject: Re: [GPSL] Earths shadow

                     

                    This is great.
                    And matches  Pretty close to what my Astronomy program shows.

                    I used the center of the Sun, If at 3 meters elevation the sun this morning rises at 7:03 AM

                    At 27,000 Meters the center of the sun is at the horizon at 6:45 IE: 18 minutes earlier

                    and at 34,000 meters the sun is centered on the horizon at 6:40  IE: 23 minutes earlier.

                    What brought this all up was we launched the balloon earlier that planned, On the ground the Sun had set. But shortly after launch the balloon suddenly lit up. It was fully illuminated by the Sun Yikes!  Didn't want that.
                    And the thought was

                    Will this eventually get back into the earths shadow? Or will the rate of rise out pace the rate the earths shadow rises. So that brought up the thought of how fast does the shadow actually rise?

                    I am willing to bet it matters on day of year, and the locations latitude also.  But sadly my math skills are no where up for this calculation.

                    I wonder if it is even linear. I'm imagining in my mind that it is not. I'm thinking is rises slower when it is low, and much faster as it gets higher.


                    Anyone?

                    Joe WB9SBD


                    The Original Rolling Ball Clock
                    Idle Tyme
                    Idle-Tyme.com
                    http://www.idle-tyme.com

                    On 2/9/2013 8:49 PM, L. Paul Verhage wrote:

                    Here's now I figured when to launch a balloon to see sunrise. Make adujstments to the numbers to suit your particular situation.

                    Using the equation distance (miles) = square root [1.5 * altitude (feet)], the horizon is 387 miles at an altitude of 100,000 feet. Earth rotates once every 24 hours and Earth's circumference at a latitude of 40 degrees is 19,151 miles (25,000 miles * cos(40 deg)). So Earth rotates about 798 miles per hour at 40 deg N. If the near spacecraft sees the horizon 387 miles away, it will observe sunrise 29 minutes earlier than the ground below it.

                    I figure the balloon will take 90 minutes to get to 100,000 feet and I want the balloon there 29 minutes before local sunrise. So the launch must occur 1 hour 59 minutes prior to local sunrise.    

                    I have three flights that observed sunrise in near space. However, I think it would be more interesting to dawn in near space. So my next attempt will try to observe the horizon perhaps 30 minutes prior to sunrise.

                    Paul

                    On Fri, Feb 8, 2013 at 6:54 PM, Joe <nss@...> wrote:

                    agreed!


                    The Original Rolling Ball Clock
                    Idle Tyme
                    Idle-Tyme.com
                    http://www.idle-tyme.com

                    On 2/8/2013 7:18 PM, Mark Conner wrote:

                    It wasn't me, and probably by the time I sort out all the math I'm pretty sure the balloon will be in darkness.

                     

                    This would have been interesting to see.  On the pre-sunrise flights I've participated in, the balloon becomes brilliant very quickly as you're going up into daylight as sunrise is "moving down" towards the surface, so you have a rapid transition from darkness to full illumination.  For a post-sunset flight, you're chasing the earth's shadow moving up in the atmosphere, overtaking it and going into the light, then as the angles get more shallow the shadow will move up more rapidly and overtake the balloon again.  The math on this could get pretty complex.

                     

                    73 de Mark N9XTN

                    On Fri, Feb 8, 2013 at 6:39 PM, James Ewen <ve6srv@...> wrote:

                    On Fri, Feb 8, 2013 at 4:43 PM, Joe <nss@...> wrote:

                    That again will be a factor of latitude and time of year, and also the
                    altitude that you are interested in. Probably the term you are looking
                    for is local sunset at a specific altitude.

                    Some of the smart guys here know the formula... I think Mark Conner
                    was posting some information like that a while ago.

                    --
                    James
                    VE6SRV

                     




                    --
                    Onwards and Upwards,
                    Paul

                     

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