6153 replies to this topic

[Hazzard]

skyeagle409, on 29 April 2011 - 08:53 PM, said:

Apparently, that is not true and the skeptics need to learn lessons,  and not learn them the hard way , as has been the case more than 40 times as listed where they said this, and said that, and later, learned the lesson the hard way when the rest of the story was later revealed, but I still think it is unfair for you to  think that I was just playing games by withholding certain information as I don't see it as my fault for  the way the skeptics were responding before I  pulled out the other book.

Bla, bla, bla, bla, bla...

To tell you the truth, Skyeagle, I dont "enjoy" your hand waiving/song and dance routine as much as I used to.

In fact, I find your hysterical UFO mantra rather dull these days, and quite embarrasing.

Edited by Hazzard, 29 April 2011 - 10:34 PM.

[booNyzarC]

skyeagle409, on 29 April 2011 - 08:47 PM, said:

No, I understand what was said about ditching the flares before landing, but, that close to DMAFB, they would not be ditching at such a high altitude. One another note, the Air Force first denied any involvement, and then, said that an A-10, was responsible, and now, it was a flight of A-10s that were ditching flares.

Finally you at least acknowledge that you understand the notion that the flares could have been jettisoned.  Thank you for that skyeagle, much appreciated.  I actually consider that progress.

Regarding the inconsistencies, it is fairly simple.  The initial people asked simply didn't know because the A-10's that ditched the flares were based out of Tuscon, not out of Luke Air Force Base.  The people initially asked responded as expected, by describing operations run from Luke.  There is no conspiracy here, it is pretty simple really.

skyeagle409, on 29 April 2011 - 08:47 PM, said:

Well, it is not practical to ditch flares behind one another at night due to the fac t that interferes with the pilot's night vision and, would not have been ditching flares that night in such a manner, and, there is more to that story as well, which I have already brought up. So what we have here is, the initial denial of the Air Force, and then, a single A-10 dropping flares on a mission, which would have been below 6000 feet,  and now, a flight of A-10s ditching flares. Look at that video again. That is not a flight of A-10s ditching flares at 10 PM, and there was another reason as well that I have brought up in that regard.

This argument doesn't make any sense.  Ditching flares at night interferes with the pilot's night vision?  You are joking right?  When flares are deployed at night for legitimate use are you saying it doesn't interfere with their night vision?  Do you see how this argument doesn't make sense?

I explained the initial confusion regarding the conflicting answers provided when Luke was originally asked.  Very simple.

skyeagle409, on 29 April 2011 - 08:47 PM, said:

The Air Force now has multiple explanations for the "Phoenix Lights" and even a denial of involvement for what had happened that night. It is clearly evident that Air Force is covering up the Phoenix sightings. On another note, and looking at that video again, at 50 miles away, what is the distance between the two outer lights, one on the extreme far left and on the other, on the far right, if they were 50 miles away. There will be some time / distance calculations in regards to what I have said, later.

No cover-up.  Very simple reasons for the initial confusion.

As for the distance between flare 1 and flare 2 has been estimated to be about 7.5 miles.  I'm working on a recalculation over the weekend hopefully to confirm or more precisely identify this distance, but as far as rough estimates go it is probably fairly accurate.

skyeagle409, on 29 April 2011 - 08:47 PM, said:

In regards to the side profile depiction of Phoenix and of the BGR, there is a  reason why I did not say anything about the curvature of the earth in regards to the side profile depiction of the two areas. I only provided the elevations.

Speaking of side profile depictions...  I've done a bit more with one.

Consider this image (courtesy of Google Earth) showing an elevation profile from Krzyston's house at 1640' elevation (1637' in the profile) extending about 80.3 miles at a heading of 205.26 degrees and a straight line representing line of sight above the highest point of elevation (about 4200') between him and the last flare dropped (light 9 in the Maccabee analysis).


Click Me for a Larger Version of the Same Image


This picture of the curve is grossly exaggerated by a magnitude of about 45 times, to 1.64 degrees, because the actual curvature would be visually negligible at a more accurate calculation of 0.036 degrees (and the lowest I could get it to even go with GIMP was somewhere around 0.14 degrees, virtually invisible curve).

Even with this exaggeration, at a rounded distance of 80 miles and not considering atmospheric refraction (thank you for the terminology correction bmk1245, very much appreciated! ) we can see that the lights would still be visible at an altitude of about 9700' or 9800'.  Considering refraction, light 9 would still be visible after dropping below this altitude as well.

I haven't done the math yet to compare with the visual of this sideview from Google Earth, but I doubt if it is too far off.

skyeagle409, on 29 April 2011 - 08:47 PM, said:

What this all amounts too, the Air Force did it again and mislead some people that flares were responsible for the sightings, but, that was not the case at all.

Nope, as addressed earlier, there was no cover-up.  It was just simple confusion and to be fully expected.

[booNyzarC]

lost_shaman, on 29 April 2011 - 02:12 AM, said:

Ok Sky, if you are not going to answer then the only clue to distance from the Camera to the Mountain you gave was "between" so I'll assume 25 miles.

At 25 miles (assuming the Camera is near sea level), 1,586.9 ft of this Mountain extends above the Horizon. That equates to 0.6877 degrees or 3,168.8 ft @ 50 miles. Adding this figure to the 1,652.3 ft calculated in a previous post above we get 4,821.1 ft as a minimum height for an object to be seen 50 miles away and above your hypothetical Mountain 2,000 ft tall above sea level 25 miles away in the Chicago Photo you posted.

lost_shaman, on 29 April 2011 - 05:10 AM, said:

20 Miles... = 5,991.3 ft.

I realize that this calculation is somewhat moot, because it really has no relation to the Phoenix Lights, but I thought you might appreciate these elevation details.  I tracked them down when skyeagle initially posted his picture but pretty much stopped there because I didn't see much point.

580ft elevation along the shore of Lake Michigan where the photo was snapped.
597ft elevation at the base of Willis Tower in Chicago.

[lost_shaman]

booNyzarC, on 30 April 2011 - 12:20 AM, said:

This picture of the curve is grossly exaggerated by a magnitude of about 45 times, to 1.64 degrees, because the actual curvature would be visually negligible at a more accurate calculation of 0.036 degrees (and the lowest I could get it to even go with GIMP was somewhere around 0.14 degrees, virtually invisible curve).

Even with this exaggeration, at a rounded distance of 80 miles and not considering atmospheric refraction (thank you for the terminology correction bmk1245, very much appreciated! ) we can see that the lights would still be visible at an altitude of about 9700' or 9800'.  Considering refraction, light 9 would still be visible after dropping below this altitude as well.

Hey booN,

Can you explain a bit better what you've done here? Is this just the angular calculation of altitude before a correction for the curvature of the Earth?

Thanks.

[DONTEATUS]

Oh YEah! I said,he said,She said They said,We all Are sad.
The point of the thread BE for ER visitation,Is that the Ambalance cant get here quick enough !
Help me Someone ! Sweetpumper ! My Helmut !My B.B.Q !

[Slave2Fate]

DONTEATUS, on 30 April 2011 - 02:13 AM, said:

Oh YEah! I said,he said,She said They said,We all Are sad.
The point of the thread BE for ER visitation,Is that the Ambalance cant get here quick enough !
Help me Someone ! Sweetpumper ! My Helmut !My B.B.Q !

Are you saying the ETH needs some CPR by an EMT otherwise it'll be KIA?

[mcrom901]

lost_shaman, on 30 April 2011 - 01:17 AM, said:

Can you explain a bit better what you've done here?

let's see what we've got here....

[booNyzarC]

lost_shaman, on 30 April 2011 - 01:17 AM, said:

Hey booN,

Can you explain a bit better what you've done here? Is this just the angular calculation of altitude before a correction for the curvature of the Earth?

Thanks.

No problem lost_shaman, I had actually been planning to do more with it but when skyeagle brought up the side view point again I decided to just drop that in.  This will probably be more information than you're looking for, but oh well, ignore what you don't care about.

What I did is identify the actual location where Mike Krzyston filmed his footage.  I thought this would be necessary because when you plug his address into Google Earth it centers on his house, but not the exact spot he filmed from on his patio.  The possible error of this general location became apparent when I reviewed my initial confirmation of Maccabee's numbers and noticed that the line from K's assumed vantage point to my re-created triangulation of light 9 in the array actually crossed slightly to the left of Hayes Peak, as you can see here:

 incorrect line k to 9.jpg   80.54K   6 downloads

This made me question my placement of light 9 from the array and my reliance on the assumed positioning of the camera's point of view simply from Google Earth's automatic placement at Latitude: 33°36'46.55"N Longitude: 112° 5'24.13"W.  To really be as accurate as possible, I needed to know about where he had filmed the original footage.  After reviewing a lot of material I narrowed it down to his patio, which is actually pretty large, meaning he could have filmed from anywhere along the south edge of his house and the point of view differences could have an impact on reproducing Maccabee's numbers for confirmation.

I finally found reference in the daytime footage collected for the analysis that Cognitech had done.  Not sure why I didn't go there first to look, but oh well...  I had been so focused on the ridge of the mountains when I grabbed these screen captures that I totally missed the pool in the foreground.  (Yet more evidence that human beings typically make for bad eye-witnesses! LOL)

 k position day shot 1.jpg   86.55K   8 downloads

After repositioning the camera, I then lined up the angle to match with the vanishing point of light 9 against the ridge to determine the azimuth as roughly 205.26 degrees (which Maccabee had listed as 205 +/- 1 degree).  I extended that line of sight for 80 miles from K's patio to discover that I had indeed placed light 9 incorrectly after all.  While I was there I decided to take an elevation profile which is a feature of Google Earth when you have a defined Path.  That profile is what you see at the bottom of my linked picture.



I copied the scale from that section and extended it upward because the scale stopped at the high-point in the profiled path, which was from the western slope of Hayes Peak.  Each vertical section represents an elevation change of 500 feet and each horizontal section represents a distance of 7.5 miles.  I used GIMP for this (and most of my image editing, because it is free and I'm too cheap to buy Photoshop and refuse to pirate software as a general rule) and applied the upper layer scale with about 50% transparency.

From this baseline image I wanted to find the general curvature of the earth in order to apply that curve to the scaled elevation profile.  I found a reference to this curvature as about 16' 8" every 5 miles, extended this by half of the distance covered in this image to 40 miles, or 133.33 feet which came to an angle of 0.03617 degrees from the center to each end.  I don't know if this is completely accurate, but I had a very busy day with work so I didn't take the time to verify it beyond that calculation.  When I tried to apply this angle of curve in GIMP I found that it wouldn't go that low for such a small image and quickly realized that it wouldn't be noticeably visible in this 80 mile cross section, so I just did a noticeably visible curve instead and measured the angle afterward to find it was about 1.64 degrees.

After applying the curve effect to the image with GIMP I drew a straight line from K's point of view to the edge of upper scale and found it hit in the upper section between 9500' and 10000'.  My eyeball estimate placed it at around 9700' or 9800' so I ran with it.

Hope that helps clarify and doesn't boor everyone to tears.  I wanted to be more detailed, but as I said, I was pretty busy with work today.

From here I plan to more accurately triangulate the actual positions of each flare as it appears and as it disappears using the K, L, and R videos primarily and then confirm against King's video.  It will probably take me a while because I am easily distracted.

[booNyzarC]

Speaking of analysis of the Phoenix Lights... go check out what AlienDan just put together...

I'm impressed.

[lost_shaman]

booNyzarC, on 30 April 2011 - 03:06 AM, said:

After repositioning the camera, I then lined up the angle to match with the vanishing point of light 9 against the ridge to determine the azimuth as roughly 205.26 degrees (which Maccabee had listed as 205 +/- 1 degree).  I extended that line of sight for 80 miles from K's patio to discover that I had indeed placed light 9 incorrectly after all.  While I was there I decided to take an elevation profile which is a feature of Google Earth when you have a defined Path.  That profile is what you see at the bottom of my linked picture.



I copied the scale from that section and extended it upward because the scale stopped at the high-point in the profiled path, which was from the western slope of Hayes Peak.  Each vertical section represents an elevation change of 500 feet and each horizontal section represents a distance of 7.5 miles.  I used GIMP for this (and most of my image editing, because it is free and I'm too cheap to buy Photoshop and refuse to pirate software as a general rule) and applied the upper layer scale with about 50% transparency.

Hey booN,

I follow up until this point where I think you've done fine work showing the Tangent line (adjacent side of a right triangle) and the line of sight above the Mountain (hypotenuse side of a right triangle).

booNyzarC, on 30 April 2011 - 03:06 AM, said:

From this baseline image I wanted to find the general curvature of the earth in order to apply that curve to the scaled elevation profile.  I found a reference to this curvature as about 16' 8" every 5 miles, extended this by half of the distance covered in this image to 40 miles, or 133.33 feet which came to an angle of 0.03617 degrees from the center to each end.  I don't know if this is completely accurate, but I had a very busy day with work so I didn't take the time to verify it beyond that calculation.  When I tried to apply this angle of curve in GIMP I found that it wouldn't go that low for such a small image and quickly realized that it wouldn't be noticeably visible in this 80 mile cross section, so I just did a noticeably visible curve instead and measured the angle afterward to find it was about 1.64 degrees.

From here though you have made a serious error, I hope you don't mind it being pointed out as you said you were too busy to verify the curvature calculations.

You were correct that at 5 miles the Drop due to curvature is ~16.5 ft, however the drop itself is non-linear. i.e at 10 miles the drop is not 33 ft but rather 66 and at 40 miles it is not 133.33 ft but rather 1,057.5 ft, etc.

[booNyzarC]

lost_shaman, on 30 April 2011 - 04:21 AM, said:

Hey booN,

I follow up until this point where I think you've done fine work showing the Tangent line (adjacent side of a right triangle) and the line of sight above the Mountain (hypotenuse side of a right triangle).





From here though you have made a serious error, I hope you don't mind it being pointed out as you said you were too busy to verify the curvature calculations.

You were correct that at 5 miles the Drop due to curvature is ~16.5 ft, however the drop itself is non-linear. i.e at 10 miles the drop is not 33 ft but rather 66 and at 40 miles it is not 133.33 ft but rather 1,057.5 ft, etc.

I don't mind at all.  In fact I appreciate the education.

I looked at several different sites (briefly) to try to make a quick calculation because my interest was in making the image somewhat match with reality.  I found a reference and ran with it.

In my defense it has literally been over 22 years since I took trig in high school.  I excelled in class, but I honestly haven't used it since.  I fell back on a linear assumption, but seriously thank you for correcting me.

I'm excited on two fronts...  first, I learned (or re-learned?) something new!  And second, my graphic should turn out much better with more accurate calculations.

So I thank you for that.

Edit:

I just recalced...  can you confirm?

drop in curvature at 40 miles would be 1057.5'
side a = 1057.5'
side b = 211200'
side c = *211202.64749'
Angle A = 0.28688 degrees

That would mean that my image is still grossly overstated?  Not by a factor of 45, but rather at a factor of about 5.7 (1.64/0.28688)?

Edited by booNyzarC, 30 April 2011 - 04:52 AM.

[lost_shaman]

booNyzarC, on 30 April 2011 - 04:38 AM, said:

I don't mind at all.  In fact I appreciate the education.

I looked at several different sites (briefly) to try to make a quick calculation because my interest was in making the image somewhat match with reality.  I found a reference and ran with it.

In my defense it has literally been over 22 years since I took trig in high school.  I excelled in class, but I honestly haven't used it since.  I fell back on a linear assumption, but seriously thank you for correcting me.

I'm excited on two fronts...  first, I learned (or re-learned?) something new!  And second, my graphic should turn out much better with more accurate calculations.

So I thank you for that.

Hey booN,

No Problem. This formula (distance in Miles squared, divided by 1.513) will work briliantly to determine the drop. It's very simple.

BTW, if it's any consolation I've learned this all from scratch myself.

[mcrom901]

lost_shaman, on 30 April 2011 - 05:03 AM, said:

BTW, if it's any consolation I've learned this all from scratch myself.

ah... love that word... "scratch"



time for a sagan interlude....

“If you want to make an apple pie from scratch, you must first create the universe.”

[booNyzarC]

lost_shaman, on 30 April 2011 - 05:03 AM, said:

Hey booN,

No Problem. This formula (distance in Miles squared, divided by 1.513) will work briliantly to determine the drop. It's very simple.

BTW, if it's any consolation I've learned this all from scratch myself.

Again, much appreciated.  First, can you confirm my conclusions above ( I edited my post, apologies for that. )

Second, it took going through this process for me to fully understand the usage of your calculations.  I've been meaning to come back and figure it out, but now I understand the application pretty well I think.  So again, thank you.

Third, may I ask how you arrived at 1.513 as the proper divisor for curvature calculations?  I'm not doubting, I just want to more fully understand the reasoning behind his particular number.

[lost_shaman]

booNyzarC, on 30 April 2011 - 04:38 AM, said:

Edit:

I just recalced...  can you confirm?

drop in curvature at 40 miles would be 1057.5'
side a = 1057.5'
side b = 211200'
side c = *211202.64749'
Angle A = 0.28688 degrees

That would mean that my image is still grossly overstated?  Not by a factor of 45, but rather at a factor of about 5.7 (1.64/0.28688)?

Hey booN,

There is no need to calculate another triangle here as the curvature of the Earth is a correction that gets added to the first triangle calculation. Take K's video as an example. K is ~1,600 ft ASL so we subtract that from the mountain height (4,300 ft) = 2,700 ft. We could further correct for curvature here if we wanted to @ 26 miles, but at anyrate this is the rough Tangent line (adjacent side of a right triangle) the above figure (2,700 ft) is the 'a' side (opposite side of a right triangle).

Now the whole point of the above is simply to try to answer the question, what Angle in degrees do the Mountains subtend above K's horizontal tangent (straight horizontal line)?

It is this Angle in degrees that can then be calculated out to distance. i.e. a new triangle with the same Angle and the new distance (adjacent side of new triangle) which is all we need to calculate the new opposite side of the triangle (Altitude of an object). Now we simply must add the correction for Earth's curvature to the Altitude figure we just calculated.

Do you get what I'm saying? We only need to calculate a triangle to understand an Obstruction (such as a Mountain) that impedes the view of the horizon. Without such an Obstruction we are simply calculating the curvature of the Earth. i.e. there is no need for a triangle calculation at all.

The Earth's curvature is basically constant but non-linear at distance.

Edited by lost_shaman, 30 April 2011 - 06:55 AM.