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discrepancies in distance perspectives

• DaddysKat 3,554 Posts

  Fri, Jul 26 2013 6:29 PM

  andyson:

  Oh so close DaddyKat!  But not within 2 degrees, sorry.  Next try, don't guess at the length of the shadow, look at the photo again and with a couple of estimates you can calculate the length.

  So your estimate of the length of the flag's shadow is exact?  This gal's calling BS,  professor!  LOL.  

  Since the rules of perspective (in art) are at play, and since the flag is not at the true corner of the grid or intersects the grid at the true corner, the difference in length between the shadow and each corner of the grid is merely speculative ... and while your guess is different than mine (and probably closer, since you employed the Pythagorean theorem), you can't honestly tell me that you know the shadow's true length.  (How's that for a whiny student's argument?  Pretty slick employing the laws of perspective, IMO!) 

  Anyway, thanks for the "so close" ... I'm going to leave that shadow length to the real math majors and engineers on the site to determine the angle of the sun.

  Do I get kudos for using the correct calculation at least?  I'll take a C- since it's just a quiz!  I did show my work (somewhat)!

  This little nerd thanks you for the fun, professor Andyson!  

  Excuse me ... I think the 28,42 degree angle of the sun says it's Jack:45!

  Edit:  6' shadow = 49 degrees.  ROFL!

   

• andyson 6,415 Posts

  Fri, Jul 26 2013 9:32 PM

  DaddysKat:

  Edit:  6' shadow = 49 degrees.  ROFL!

  

  Well done DaddysKat!!  Glad you had fun with it!

• DaddysKat 3,554 Posts

  Fri, Jul 26 2013 9:38 PM

  andyson:

  Well done DaddysKat!!  Glad you had fun with it!

  Thanks, professor Andyson!

  Kat

• DaddysKat 3,554 Posts

  Sat, Jul 27 2013 10:08 AM

  Hey Mr. Math (aka professor Andyson) ...

  I'm sick and tired of trying to find your "X".  Just accept the fact that she's gone and move on dude!  lol ....

  have a nice weekend!

  Kat

   

• Devius1 7 Posts

  Sat, Jul 27 2013 11:02 PM

  bahahahaha , i love it :) you obviously have alot more time to sit on the net than me though. I happen to live in a part of the country that half the people or more speak spanish and every one under  ohh idk 40 writes in text so the english laguage is butchered beyond recognition.  I will except the FACT you are an engineer simply because only an engineer could in the same paragraphy so grossly simplify AND over complicate the question. it  is a loaded question and cant be truley calculated with the information at hand and you know that however.............. 51.5 degrees give or take , or aprx 3:15 to 3:30 in the afternoon, just off the top of my head, minimal math and no paper riquired. Making terrible assumtions to fill in the gaps in your question of course.

• Devius1 7 Posts

  Sat, Jul 27 2013 11:29 PM

  I will explain my self and you will have to look at things from a different perspective why i said what i said. I am an intuititive mathmatician so i go about things entirely backwards. I came up in the field doing structural iron work and cwertified welding and didnt get a degree till 20 years later so mine tends to b a more hands on practicle aproach

       Here goes: ASSUMPTIONS

  given a true 12hr daylight, ie the equinox

  given that we are on or near the equator which for me isnt far from the truth

  given that there is 0 relative humidity and particle content to the air

  givent that we are dealing witha flat level plane FALSE

  and given that the flag stick represents a true 90 degrees to the plane

  the logic then is as follows

   

  first off throw the units of measure out the window they are irrelevant to this situation,  lets say they are cm  just for convieniance. mentaly draw a protractor and knowing what we know about triangles then 3:00 pm would be 45 degrees roughly  and the shadow would b the same length as the pin because equalangular means equaldistant and vice versa. So understanding that all equsions must balance on both sides and making a rough estimate of that shadow length at 6 ft based on the fact that the hypoyinuse of a 2 foot equidistant right triangle is aprx 2.75 ( rounding off here  lets face it we arent launching a space shuttle)  that shadow is aprx 1/7th shorter than the pin so if we chage the distances by 1/7 th then the angles change by the same ratio. Once again rounding for expediance 1/7th adjustment offf of that 45 is aprx 6.5 degrees, and there is your 51.5 degrees or between 3:15 and 3:30 given the above assumptions :) its late i am drunk lol but should b close enough.

  PS: have fun correcting all this spelling while i am out on the gulf of mexico fishing again today lol my fingers r near nerve dead i cant type for shite n dont care enough to correct

   

   

• andyson 6,415 Posts

  Sun, Jul 28 2013 6:16 AM

  Devius1:

  there is your 51.5 degrees or between 3:15 and 3:30 given the above assumptions :) its late i am drunk lol but should b close enough.

  Not a bad SWAG*!   But to get back on topic, where on WGT was it you saw the 45 ft flagstick shadow that started this whole discussion?

• IH8kidz 907 Posts

  Sun, Jul 28 2013 9:16 AM

  Why does popcorn pop? Who cares!

• DaddysKat 3,554 Posts

  Sun, Jul 28 2013 11:03 AM

  Devius1:

  given a true 12hr daylight, ie the equinox

  given that we are on or near the equator which for me isnt far from the truth

  given that there is 0 relative humidity and particle content to the air

  givent that we are dealing witha flat level plane FALSE

  and given that the flag stick represents a true 90 degrees to the plane

  the logic then is as follows

  Sorry Devius1, the snapshot was taken from Bethpage Black ... Lat: 40.73, Long: -73.44 (just a tad north of the equator).  Assuming fall equinox (Sep 20), Sunrise from these coordinates would be 6:41 and Sunset would be 6:55.

  Devius1:

  between 3:15 and 3:30 given the above assumptions

  In order to determine the time from shadow length, wouldn't you need to know the direction of true north?  

  So ... does anybody really know what time it is ..... does anybody care (I don't know) ... (lyrics to an old Chicago tune?)

• DaddysKat 3,554 Posts

  Sun, Jul 28 2013 12:30 PM

  Here's one for all you engineers and math majors: