Geek Break: Calculating Elevation of Jet Transiting the Sun

Brandon Ghany on Flickr

After my last post about capturing a jet flying in front of the Sun, I saw this image Brandon Ghany posted on Flickr of a jet crossing in front of the Moon and what caught my attention is how much smaller the jet is than in my video. 

From Brandon's description, this jet was flying at about 12,500 feet. Clearly, the jet I caught was flying much lower. But how low? I was curious enough to try figuring it out. 

The first factor to consider is that the Sun and Moon are similar but not exactly the same size from our viewpoint. Actually, the distance to the Moon varies a little at different points in its orbit so sometimes it is closer and thus a bit larger than the Sun (necessary to have a total solar eclipse) and sometimes it is farther away and thus a bit smaller than the Sun (leading to views like the Ring of Fire of an annular eclipse such as the one on October 14th this year). But in general terms, they are both roughly a half a degree across viewed from Earth. That means if Brandon had captured a jet at 12,500 feet passing in front of the Sun instead of the Moon, his image would have looked similar, at least in terms of the relationship between the size of the jet and the size of the celestial orb it passed in front of. 

The second factor is that while I know the angular size of the sun (reported by Stellarium as 31.68 arc minutes at the time of my video), for me to determine things like the jet's altitude, I need to know the jet's angular dimension, too. 

To calculate this, I captured a single frame from my video, cropped and rotated it then used drafting software (FreeCAD) to measure the relative difference between the width of the jet's fuselage and the width of the Sun. The ratio came out to 55.25% which means the width of the jet body is 17.5 arc minutes. 

So, how did you do in high school or college trig class? I muddled through but that was also so long ago, I just don't recall the exact formula to use. Thanks to Google, though, it was pretty easy to find what I needed.

d = ( w / θ ) * 60

Simple, right? Well, except I'm missing one variable: w, or width. Google to the rescue again. If I assume that this jet was something like a Boing 737 or an Airbus 320, both pretty common models used by airline companies flying out of DFW and Love Field, then the width of the fuselage is approximately 13 feet. Plugging that into the formula as w and converting arc seconds to degrees and using that as θ, that tells me that the jet was approximately 2,674 feet away when it passed over our neighborhood. 

But was that how high it was flying? Not exactly since it wasn't directly over my backyard. Again checking Stellarium, I found that the Sun's altitude (its angle above the horizon) was about 62.65 degrees.

Dusting off more trigonometry, that gives me two variables of a right triangle, the angle and the hypotenuse. From those, I can calculate the jet's elevation (the opposite side of the triangle) and the distance over ground to the jet's position when it transited the Sun. 

o = h * sin(θ)

a = h * cosin(θ)

Solving these is pretty straightforward with a modern calculator or using spreadsheet software like Google Sheets. Note that these formulas require the angle to be expressed in radians but spreadsheets have a function for that, too, so you don't have to remember the formula for converting degrees to radians:

θ radians = θ degrees * π / 180

So what's the answer? Based on the numbers above, at the time it passed in front of the Sun that jet was flying at an elevation of approximately 2,377 feet and its position over land was about 1,225 feet from my backyard, about a quarter of a mile to the southwest.

I would have thought the jets flying over East Dallas would be higher than that but the data and formulas say it has to be somewhere between 2,000 and 3,000 feet up. To see for yourself, make a copy of my spreadsheet and play around with what happens when you change the jet fuselage width (cell B7). Even if you change it to the size of a jumbo 747, 21.3 feet, the calculated elevation of the jet is still well under 4,000 feet.

Another interesting exercise would be to play around with the ratio of the jet fuselage width to the orb size to see what it has to be for the jet in Brandon's image to be flying at 12,500. Give it a try.

Remind you of one of those math word problems you hated solving in school?

Yeah, but pretty cool, huh?

Note

It's been decades since I was in school so if you find that I've made a mistake in my calculations, I'd be happy to hear about it in the comments.