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psylant
World Explorer


Reged: 07/23/05
Posts: 367
Loc: Calgary, AB, Canada
Re: math can be so useful? [Re: AZTraveller]
      #81129 - 08/23/05 12:36 PM

Quote:

If it helps this image was captured January 16th 2003 at exactly 10:18 am. Does that help. The plane and its shadow from 450 kilometers in space will be the same size as measured on the ground. So you can use the back shadow and the white plane and see how high it is.



How do you locate that info? I was just calculating the sun's azimuth based the shadow's position, then using the date from the DG layer that lined up with the photo in GE. Is that information accurate? Or is the photo's time and date recorded somewhere else? There's no DG layer from Jan 16 2003, but there is one from Jan 16 2005, but that's completely covered by cloud. Where is the time of the photo? Where's the date of the photo you're using? Time is not a factor with the sun calculator I used, but azimuth is and that's easy to calculate directly.

Thanks!

Edited by psylant (08/23/05 01:11 PM)


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AZTraveller
Master Cartographer


Reged: 07/15/05
Posts: 1388
Loc: Arizona
Re: math can be so useful? [Re: psylant]
      #81445 - 08/23/05 05:23 PM

No magic use the DG layers to find an exact fit recatangle and the newest date is the date the image was captured. The time is also easy. The satellite orbits every 93.5 minutes and crosses the equator at 10:30 am solar time every orbit (dark side 10:30 pm). The satellite travels North to South .....you take the lattitude and find what fraction of 90 degrees and then mutiply by a quarter of 93.5. If you are North of the equator subtract and add if you are South.

Youhave to know which time zone you are in relative to Greenwich to adjust for any local time anomolies. Summer time winter time and whether your local time follows the rules.

Us cowboys in Arizona never change our clocks and I know that happens other places

When I crack the time shift for the blue images we will be able tto calculate altitude and speed.


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psylant
World Explorer


Reged: 07/23/05
Posts: 367
Loc: Calgary, AB, Canada
Re: math can be so useful? [Re: AZTraveller]
      #81527 - 08/23/05 06:37 PM

Quote:

No magic use the DG layers to find an exact fit recatangle and the newest date is the date the image was captured. The time is also easy. The satellite orbits every 93.5 minutes and crosses the equator at 10:30 am solar time every orbit (dark side 10:30 pm). The satellite travels North to South .....you take the lattitude and find what fraction of 90 degrees and then mutiply by a quarter of 93.5. If you are North of the equator subtract and add if you are South.




I'm still confused. If the latest date is the image used, then what's with this? Clicking on the icon for the 2005 DG layer over the C-17 at Grant County, and it's completely overcast. I find the March 8 2004 image corresponds more accurately with the one in GE, compared with any other dg layer. Feb 11 2003 comes close, but those fields southeast of the airport don't match up, and it's not cropped in the same area as March 8.

The satellite's orbit info is certainly useful; that'll come in handy!

Cheers.


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aspeedbump
Tourist


Reged: 08/20/05
Posts: 11
Re: Who would've thought math could be so useful? [Re: psylant]
      #81886 - 08/24/05 04:02 AM

Yes, yes... after making my post I decided to re-draw the line, and realized that he was, in fact, lined up for the runway... my mistake.
Interesting caculations getting the altitude... thanks.


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dyatsko
Tourist


Reged: 09/03/05
Posts: 54
Re: Who would've thought math could be so useful? [Re: psylant]
      #201657 - 11/17/05 03:08 PM

Reference post of 08/22/2005.

A standard approach 'glideslope' is 3 degrees.
If the C-17 is 'on the glideslope', calculate it's altitude the following way.

Measure the distance from the aircraft to the touchpoint in nautical miles (in this case approx 1.1 NM).
By high school triginometry:
tan(3 degrees) x ft per nautical mile will give you the glide slope altitude at 1 NM.
tan(3 degrees) = 0.0524
x feet in a nautical mile (6076.11549) equals
318.3885 times 1.1equals
350.2274 or rounded to 350 feet.
Therefore for every nautical mile from the touchdown point the proper glideslope altitude is 318.3885 feet

If the C-17 is on the glideslope it's alltutude is 350 ft


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akav8r
First Post


Reged: 06/30/07
Posts: 1
Re: Who would've thought math could be so useful? [Re: dyatsko]
      #935401 - 06/30/07 10:00 AM

To complicate things, the C-17 typically flies assualt approaches at up to 5 degrees. At one mile from the runway, they are often 500' in the air.

This is the practice assault landing runway for McChord AFB C-17's, and is only 3500 feet long. The black rubber buildup comes from application of full braking pressure immediately after touchdown. Try that in an airliner! We often do a full-stop landing, but quite often will fly a "GOAT," or "go-around after touchdown." It's a very brief touch-and-go after landing inside the "zone" or first 500' of the runway. Pretty fun to do.


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Davidgw
Tourist


Reged: 10/06/07
Posts: 2
Loc: Covington, WA, King County
Re: Who would've thought math could be so useful? [Re: akav8r]
      #1021861 - 10/07/07 05:34 PM

There is an airdrop practice zone in that area. Check out this geocache listing: http://www.geocaching.com/seek/cache_details.aspx?guid=30392bd2-0f0f-4e88-8f8d-a67bacd3291d

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