Tämä nyt ei hirveästi varmaankaan kiinnosta asiaan perehtymättömiä. Mutta viime vuonna B. Borg peräänkuulutti ns. Kustavin ufokuvan tutkimustuloksia erään kanadalaisen kaverin toimesta. Julkaisen tässä pätkää sähköpostikirjeenvaihdosta, jossa esitetään muutamia laskelmia. Poistakaa viesti, jos haluatte, mutta lähettäkää SUT:lle kopio, jos viitsitte.
From: Bruce Maccabee
To: UFO UpDates - Toronto
Cc: John Velez
Date: Wed, 5 Jan 2005 00:42:34 -0500
Subject: FWD: Finnish UFO Sighting
Thanks for the message. You wrote:
>Gentlemen,
>I did a deep-zoom - couldn't see any pixel pushing.
>Over to you.....
>Errol
-----
>Hi!
>On September 23th approximately at 5.15 pm. me amd my two
>buddies were coming back from a short fishing trip in Kustavi,
>a place located at southeastern archipelago of Finland, when
>we saw something extraordinary in the air. We soon realized
>that what we saw was some sort of a flat round metallic object,
>and that as it slowly moved in the air it made no sound at
>all.
>A friend of mine had a digital camera within reach, and he
>managed to take two pictures of the object, before it suddenly
>took off with unbeliavable pace. In just a few >seconds it
>disappeared behind the clouds.
>We send you one picture because we tought you might be
>interested. You are free to use it for your own purposes.
>For more detailed information, feel free to contact me.
I assume the picture is the full frame photo from the camera.
559 by 421 pixels.
I assume that the full width of the photo corresponds to about
40 degrees (i.e., essentially equivalent to 35 mm film and 50 mm
lens which produces width of about 39 deg wide field of view).
IF THIS IS SO, the calibration is about 40/560 = 0.071
deg/pixel.
the UO (unidentified object) is about 100 pixels wide left-to-
right (ignoring the tilt of about 7.1 degrees.
This is LARGE! 7.1 deg = approx 0.12 radians. Multiply radians
by distance to get size.
At 100 ft, 12 ft
at 1000 ft, 120 ft
at 10,000 ft 1200 ft
etc.
At ten feet, 1.2 feet; at 8 feet... hubcap or large frisbee
sized.
There appears to be noticeable blur at the left and right ends.
The amount of blur at the ends seems to be greater than at the
middle "top" and "bottom" of the image.
This could be evidence of motion smear with most of the motion
being transverse to the line of sight (left to right or v.v.)
5:15 PM... I assume this is daylight savings time. Therefore
the event occurred 4:15 PM Standard time.
The location appears to be about 21 /-1 deg east longitude by
about 60 north latitude. Sweden appears on maps to be in the
second time zone east of London. Consult http://aa.usno.navy.mil
for sun location vs time. The web site computer calculates
positions vs times. Choose 15 minute intervals for the
calculation and get the locations for a whole day.
At 1715 hours elevation of sun is 8.8 deg and 254 azimuth.
Since the sky is cloudy it may be that the sun was not visible
or only slightly visible.
The glint on the UO may be a result of reflecting bright
skylight rather than sunlight.
I am suspicious of the "reflection" in the water. Why isn't it
more round?
I wonder why he didn't send both pictures?
Anyway, analysis is incomplete without both pictures.
Since they were fishing, assume they were in a boat. Assume the
camera was H = 3 ft above the water surface. Assume the dark
feature in the water is actually a reflection of the object.
Imagine the water to be a flat mirror. Then, believe it or
not, it is possible to get a distance... IF THE TRIANGLE CLOSES
as described below.
Assume the camera is at the same height as the tops of the
distant weeds growing up out of the water. This establishes a
horizontal plane at height H above the water. Draw a straight
line from the Camera to the UO Draw a straight line from the
Camera to the Reflection. Place the reflection location so that
the angle of incidence = angle of reflection.
UO
. .
C.
---------------------------------------------------------------------------------------------horizontal
.
.
.
. .
_________________________R_______________________
The points C, R and UO should form a closed triangle.
The elevation of the UO image above the tops of the distant
weeds is about 145 pixels = 10.3 deg
The depression of the UO reflection below the tops of the
distant weeds is ALSO about 145 pix = 10.3 deg
This is also the angle between horizontal at the water surface
and a line from C to R. This is the angle of reflection.
The angle of reflection = angle of incidence. The angle of
incidence is the angle between the horizontal (water surface)
and the line from R to UO. Thus the angle between R-UO and the
horizontal equals the angle between C-UO and the horizontal
plane at height H. Thus line C-UO is parallel to line R - UO.
Ooops!! The triangle doesn't close.
However the distance to the UO can't be calculated because C-UO
and R-UO are parallel lines.
If not exactly parallel, then they are very close to parallel
meaning UO is either very far away or... this is a digital
"modification" of a scene.
Generally speaking, in a reflection situation like this the
angular elevation above the camera to the UO must be SMALLER
than the angular depression of the reflection below the camera
in order to get closed triangle and a calculatable distance to
the object.
Suppose I stretch the calculation: use the LOWEST point of the
UO to minimize the elevation angle of the line C-UO: 137 pixels
= 9.7 deg Also use the LOWEST point on the reflection to
maximize the depression angle of the line C-R: 159 pixels
= 11.3.
The height of the camera and weed tops is assumed to be 3 ft.
Hence horizontal distance from the camera to the reflection is 3
ft/tan 11.3 = 15 ft.
The radial distance from the camera to the reflection, the line
C-R, is 3 ft/sin 11.3 = 15.3 ft
Now the obtuse angle between C-R and UO -R is 180 - 2 x 11.3
= 157.4 deg. The angle between C-R and C-UO is 9.7 11.3 = 21
deg. Hence the acute angle at UO is 180 - 157.4 - 21 = 1.6 deg.
(This angle would shrink if I didn't use extremal values.)
The law of sines says that the length of line C-R divided by the
angle opposite that side, i.e., 1.6 deg, is equal to the line C-
UO divided by the sine of the obtuse angle in the triangle,
157.4 deg.
Hence line C-UO = sin(157.4) times 15.3 ft/sin(1.6) = 210 ft.
At this distance the object width would be 210 x .12 = 25 ft.
This sounds reasonable, but I am not "happy" about having to use
extremal values for the angles to the UO and the reflection in
order to get a reasonable number for distance. As one moves
away from the extremal values the acute angle at the object
shrinks and the calculated object distance increases to infinity
(as the sin of the angle approaches zero). Hmmmmm
NOTE: these are "sample calculations" which might be revised by,
for example having H be something other than 3 ft... if we know
otherwise (was he guy standing in the boat? or on a dock?)
Wonder what the other picture looks like?
From: Bruce Maccabee
To: UFO UpDates - Toronto
Cc: John Velez
Date: Wed, 5 Jan 2005 00:42:34 -0500
Subject: FWD: Finnish UFO Sighting
Thanks for the message. You wrote:
>Gentlemen,
>I did a deep-zoom - couldn't see any pixel pushing.
>Over to you.....
>Errol
-----
>Hi!
>On September 23th approximately at 5.15 pm. me amd my two
>buddies were coming back from a short fishing trip in Kustavi,
>a place located at southeastern archipelago of Finland, when
>we saw something extraordinary in the air. We soon realized
>that what we saw was some sort of a flat round metallic object,
>and that as it slowly moved in the air it made no sound at
>all.
>A friend of mine had a digital camera within reach, and he
>managed to take two pictures of the object, before it suddenly
>took off with unbeliavable pace. In just a few >seconds it
>disappeared behind the clouds.
>We send you one picture because we tought you might be
>interested. You are free to use it for your own purposes.
>For more detailed information, feel free to contact me.
I assume the picture is the full frame photo from the camera.
559 by 421 pixels.
I assume that the full width of the photo corresponds to about
40 degrees (i.e., essentially equivalent to 35 mm film and 50 mm
lens which produces width of about 39 deg wide field of view).
IF THIS IS SO, the calibration is about 40/560 = 0.071
deg/pixel.
the UO (unidentified object) is about 100 pixels wide left-to-
right (ignoring the tilt of about 7.1 degrees.
This is LARGE! 7.1 deg = approx 0.12 radians. Multiply radians
by distance to get size.
At 100 ft, 12 ft
at 1000 ft, 120 ft
at 10,000 ft 1200 ft
etc.
At ten feet, 1.2 feet; at 8 feet... hubcap or large frisbee
sized.
There appears to be noticeable blur at the left and right ends.
The amount of blur at the ends seems to be greater than at the
middle "top" and "bottom" of the image.
This could be evidence of motion smear with most of the motion
being transverse to the line of sight (left to right or v.v.)
5:15 PM... I assume this is daylight savings time. Therefore
the event occurred 4:15 PM Standard time.
The location appears to be about 21 /-1 deg east longitude by
about 60 north latitude. Sweden appears on maps to be in the
second time zone east of London. Consult http://aa.usno.navy.mil
for sun location vs time. The web site computer calculates
positions vs times. Choose 15 minute intervals for the
calculation and get the locations for a whole day.
At 1715 hours elevation of sun is 8.8 deg and 254 azimuth.
Since the sky is cloudy it may be that the sun was not visible
or only slightly visible.
The glint on the UO may be a result of reflecting bright
skylight rather than sunlight.
I am suspicious of the "reflection" in the water. Why isn't it
more round?
I wonder why he didn't send both pictures?
Anyway, analysis is incomplete without both pictures.
Since they were fishing, assume they were in a boat. Assume the
camera was H = 3 ft above the water surface. Assume the dark
feature in the water is actually a reflection of the object.
Imagine the water to be a flat mirror. Then, believe it or
not, it is possible to get a distance... IF THE TRIANGLE CLOSES
as described below.
Assume the camera is at the same height as the tops of the
distant weeds growing up out of the water. This establishes a
horizontal plane at height H above the water. Draw a straight
line from the Camera to the UO Draw a straight line from the
Camera to the Reflection. Place the reflection location so that
the angle of incidence = angle of reflection.
UO
. .
C.
---------------------------------------------------------------------------------------------horizontal
.
.
.
. .
_________________________R_______________________
The points C, R and UO should form a closed triangle.
The elevation of the UO image above the tops of the distant
weeds is about 145 pixels = 10.3 deg
The depression of the UO reflection below the tops of the
distant weeds is ALSO about 145 pix = 10.3 deg
This is also the angle between horizontal at the water surface
and a line from C to R. This is the angle of reflection.
The angle of reflection = angle of incidence. The angle of
incidence is the angle between the horizontal (water surface)
and the line from R to UO. Thus the angle between R-UO and the
horizontal equals the angle between C-UO and the horizontal
plane at height H. Thus line C-UO is parallel to line R - UO.
Ooops!! The triangle doesn't close.
However the distance to the UO can't be calculated because C-UO
and R-UO are parallel lines.
If not exactly parallel, then they are very close to parallel
meaning UO is either very far away or... this is a digital
"modification" of a scene.
Generally speaking, in a reflection situation like this the
angular elevation above the camera to the UO must be SMALLER
than the angular depression of the reflection below the camera
in order to get closed triangle and a calculatable distance to
the object.
Suppose I stretch the calculation: use the LOWEST point of the
UO to minimize the elevation angle of the line C-UO: 137 pixels
= 9.7 deg Also use the LOWEST point on the reflection to
maximize the depression angle of the line C-R: 159 pixels
= 11.3.
The height of the camera and weed tops is assumed to be 3 ft.
Hence horizontal distance from the camera to the reflection is 3
ft/tan 11.3 = 15 ft.
The radial distance from the camera to the reflection, the line
C-R, is 3 ft/sin 11.3 = 15.3 ft
Now the obtuse angle between C-R and UO -R is 180 - 2 x 11.3
= 157.4 deg. The angle between C-R and C-UO is 9.7 11.3 = 21
deg. Hence the acute angle at UO is 180 - 157.4 - 21 = 1.6 deg.
(This angle would shrink if I didn't use extremal values.)
The law of sines says that the length of line C-R divided by the
angle opposite that side, i.e., 1.6 deg, is equal to the line C-
UO divided by the sine of the obtuse angle in the triangle,
157.4 deg.
Hence line C-UO = sin(157.4) times 15.3 ft/sin(1.6) = 210 ft.
At this distance the object width would be 210 x .12 = 25 ft.
This sounds reasonable, but I am not "happy" about having to use
extremal values for the angles to the UO and the reflection in
order to get a reasonable number for distance. As one moves
away from the extremal values the acute angle at the object
shrinks and the calculated object distance increases to infinity
(as the sin of the angle approaches zero). Hmmmmm
NOTE: these are "sample calculations" which might be revised by,
for example having H be something other than 3 ft... if we know
otherwise (was he guy standing in the boat? or on a dock?)
Wonder what the other picture looks like?
Vastaukset 3
Vanhimmat
Mutta tuo teksti ON JO kirjassani ' Outoja kohteita Suomen taivaalla ' ja on ollut siellä jo kohta vuoden.
Tuossa kirjassani olen myös todennut että tuo nimetän kanadalainen tutkija on tunnistamatta ja että hän erehtyy ajassa ja käyttää laskelmissaan Ruotsin aikaa Suomen ajan sijasta.
on sama kuin alkuperäinen Antti A. niin asiaa auttaisi kovasti jos saisin tietää Antti A: n oikean nimen, osoitteen jne.
Oikean nimen voi kertoa s- postitse:
b.borg ( at ) kolumbus.fi
Täytyy neuvotella kalatoverin kanssa. Voi olla, että palaan myöhemmin asiaan.