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Hazzard

Best evidence for ET visitation - 3rd edition

6,154 posts in this topic

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 ! :rolleyes:

Help me Someone ! Sweetpumper ! My Helmut !My B.B.Q ! :wacko:

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

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

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

Math-Jokes-math-20787066-550-400.jpg

:rolleyes:

22698256v4_225x225_Front.jpg

:devil:

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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:

post-105506-0-28417000-1304131255_thumb.

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)

post-105506-0-85944400-1304131085_thumb.

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.

5669892789_659fea7630_b.jpg

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. :hmm:

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

I'm impressed. :tu:

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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.

5669892789_659fea7630_b.jpg

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).

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.

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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. :tu:

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. :tu:

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

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

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. :tu:

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.

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

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

c010.gif

time for a sagan interlude....

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

joke.gif

:P

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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.

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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

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I'll have to look at this again in the morning. Bedtime for Bonzo.

But yes, I think that there would still be a need to determine another triangle for image manipulation purposes unless I'm mistaken, which is entirely possible.

At any rate, cheers and good night. :tu:

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I'll have to look at this again in the morning. Bedtime for Bonzo.

But yes, I think that there would still be a need to determine another triangle for image manipulation purposes unless I'm mistaken, which is entirely possible.

At any rate, cheers and good night. :tu:

Hey booN,

We must use a triangle to calculate how an Obstruction obscures our view of the horizon. Then we must use a separate formula to calculate how far sea level drops below our view at a distance.

We can talk about this later though, it is late here on my end too.

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Hey booN,

We must use a triangle to calculate how an Obstruction obscures our view of the horizon. Then we must use a separate formula to calculate how far sea level drops below our view at a distance.

We can talk about this later though, it is late here on my end too.

You should be able to work out the formula based on the fact that one nautical mile, at sea level, subtends an angle of one minute of degree.

Circumference of the glode equal 21,600 nmi. Radius ('R') of the globe equals 3437.747 nmi.

Drop over one nmi should be R(1 - cos(1/60)).

Not very elegant, but that's just off the top of my head.

Edited by Mangoze

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alien.jpg

new equation just in

pie.jpgX sea.jpgsquare.jpged = flares.jpg

:innocent::P

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alien.jpg

new equation just in

pie.jpgX sea.jpgsquare.jpged = flares.jpg

:innocent::P

As always Im going with Bee`s equation ! ITs the Real Deal ! :wub::innocent:

post-68971-0-20208100-1304177569_thumb.j

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alien.jpg

new equation just in

pie.jpgX sea.jpgsquare.jpged = flares.jpg

:innocent::P

Yepper, Just look at that tall skyline reaching high into the sky in the distance......ah, where is my telescope?

A picture can be worth a thousand words, or should I say, numbers.

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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.

That story was made up by the AIr Force. Had the A-10s truly been dumping flares at hight atltitude, the Air Force would have acknowledge to the reporters when asked for an explanation. As it was, the Air Force denied any involvment and refused to get involved. Many weeks later, the Air Force came back and said, "oh yes, one of our A-10s was responsible for dropping flares from 6000 feet, which were ignited at 3000 feet.

When the Air Force saw that its expanation was flawed becauase there was no way that flares at 3000 feet could bee seen in Phoenix from behnid the mountain, the Air Force came back with yet another explanation that some A-10s were dropping flares from a HIGHER ALTITUDE because they can't land with flares aboard, however, that creates a bit of a problem when there were no A-10s airborned at 10 PM. Operatiing around the 10 PM time frame would have been cutting in on the Tucson folks quiet hours.

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 Air Force was responsible, so let's do a recap.

The Air Force initially denied nvolvment, and then later the Air Force had said, that the "Phoenix Lights" were flares, which were dropped from 6000 feet and ignited at 3000 feet, and then, the Air Force came back and said, that A-10s dropped flares from a high altitude.

Those lights were NOT indicative of flare drops by multiple aircraft at night. You know, there is a very good reason why pilots wear sunglasses during the day, it helps their night vision.

The people initially asked responded as expected, by describing operations run from Luke. There is no conspiracy here, it is pretty simple really.

Oh yes there was. No aircraft at Luke AFB were involved, so all the Luke PR folks Luke had to do was to make a simple phone call to DMAFB, and I am very sure they did just that. and yet many weeks later, the Air Force came back with flare drops at 6000 feet, not 15000 feet. Then, its explanation changed again when they determined their altitude figure was too low, so the Air Force had to make some adjustments to its altitude explanation and there you are, a full-blown Air Force cover-up on the level of the Roswell incident.

BTW, I happen to pass through Phoenix last night and there was no way that those lights were over the BGR.

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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.

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.

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.

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).

5669892789_659fea7630_b.jpg

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! :tu: ) 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.

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

One of the reasons I posted that photo, was to show that even the top of a 1400-foot building is barely visible above the horizon, and that with no obstacles in the foreground. Now, multiply that height 10 times, and see where the top of such a building would be in that photo.

Next, measure the distance between the two furthers lights in the Phoenix video and calculate their distances from one another. What are the distances between them if those lights were 50 miles away?

Once again, the Air Force's revised flare drop cover story was to make up for its 6000-foot error in its earlier cover story.

Edited by skyeagle409

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As response for your post #2542... Nope. No conspiracy. I already explained this to you. Go back and read my previous response.

One of the reasons I posted that photo, was to show that even the top of a 1400-foot building is barely visible above the horizon, and that with no obstacles in the foreground. Now, multiply that height 10 times, and see where the top of such a building would be in that photo.

That Chicago photo you posted is still just as irrelevant now as it was when you originally posted it.

Next, measure the distance between the two furthers lights in the Phoenix video and calculate their distances from one another. What are the distances between them if those lights were 50 miles away?

You don't pay very close attention do you? I already answered this. They were about 7.5 miles according to Maccabee's analysis.

Once again, the Air Force's revised flare drop cover story was to make up for its 6000-foot error in its earlier cover story.

Once again, there was no cover story. There were incorrect answers because the people being asked didn't have the answers. And at no point did the Air Force make a statement that a single A-10 dropped the flares at 6000 feet that night. This has been pointed out to you before...

Did you read lost_shaman's post #2257 which references the actual source of this 6,000' statement you keep on bleating about?

Here is the original source. And here is lost_shaman's post...

The actual quote... "(Our pilots) told me that at 6,000 feet and using those types of flares, you can see them from 150 miles away on a good night," Shepherd said."

http://kenny.anomalyresponse.org/MARYLAND_ANG_GUILTY.htm

Here is a question for you... Could such a statement be true?

(I say it is.)

And a bonus question... If so what was the 'clue' that you picked up on?

So you see skyeagle, this isn't an official statement of when flares were dropped that night. This is an explanation of normal usage of flares and how far away you would be able to see them.

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Taking the family on a short vacation...

Spain-Mallorca-Cathedral-Palma.jpg

When the going gets tough, hazz goes to spain. :)

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Taking the family on a short vacation...

Spain-Mallorca-Cathedral-Palma.jpg

When the going gets tough, hazz goes to spain. :)

Didn't you go to Greece just a few months ago? Must be nice to travel like that. Post some pics this time. ^_^:tu:

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Didn't you go to Greece just a few months ago? Must be nice to travel like that. Post some pics this time. ^_^:tu:

Greece was beautiful... This is where we are going this time.

http://www.molon.de/galleries/Spain/Mallorca/

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Greece was beautiful... This is where we are going this time.

http://www.molon.de/galleries/Spain/Mallorca/

As if I didn't envy you enough already! :lol:

Enjoy yourself Hazz, although I'm sure it would be tough not to in a place like that. :tu:

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Place looks beautiful. Have a great trip Hazz! :tu:

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Place looks beautiful. Have a great trip Hazz! :tu:

touché

alien.jpg

new equation just in

pie.jpgX sea.jpgsquare.jpged = flares.jpg

:innocent::P

:geek::wub::alien:

we have a winner..... prizy linky

:cat:

coming back to the flare theory... *cough* were the videos authenticated by any chance?

128799139864294582.jpg

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