(Probably incorrect- it’s been 25 years)
Get distance:
Cloud cover at 6000’; Estimated angle above horizon 45 degrees
Right triangle- short side “a†= 6000; Angle “A†opposite “aâ€= 45 degrees
“c†is the hypotenuse “b†is the long leg.
SIN A= a/c c =a/SIN A c =8485’ = distance to an object just under the cloud cover.
Get length of object:
Using the “closed fist†example the angular separation is close to 10 degrees.
Bisect the angle to get a right triangle. Angle “A†is now 5 degrees; use the previous value of “c†(distance) for the long leg in this triangle- “bâ€. “a†the short leg will be ½ the width of the craft. “c†is not needed.
Width of object;
a= b TAN A
= 742’
Width = 742’x2 = 1484’
Object at 3000’ along 45-degree line of sight. Width = 742’
Object at 1500’ Width = 371’
Object at 750’ Width, you guessed it – 185’.
Any lower than that- get the hell out of there…
Off Topic » in reply to Yah. But all that math!
Yah. YUK!
The whole thread (51 posts)
- Some hints for UFO investigators
- re: Still wondering
- Well put
- Pareidolia
- Distance
- But if you had multiple accurate and trustworthy sightings...
- "Ezekial saw a wheel, way up in the air..."