[Closed] Fans of planes and maths - a Friday morning ponderer

trigonometry needed and I have forgotten how it works sine cos tan and all that

28.3 miles. not taking into account any curvature of the earth.
(149253 feet)

Angle = 15 degrees
Height = 40,000 feet
Distance over ground = 149,282 feet or 28.27 miles

28.3 miles. not taking into account any curvature of the earth.

So exactly 28.3 miles. as the earth is flat... duh!

DrP

Ballpark it as a flat earther, x = 40,000/tan(15) = about 149,000 feet or 28 and a bit miles. Given the radius of the Earth is around 4000 miles the circumferential distance won't be much different to a chord of that length, so I'd stick at a rough answer of 28 and a bit miles.

https://globe.adsbexchange.com/

Alternatively you can just look up the plane here and get real time data. Its 2021 kids, you don't need to use your brain anymore. The internet is cleverer

Oh thank gawd. I thought this was going to contain conveyor belt theory.

I spend far too long on that site! FYI, there's a little button on the cog symbol called 'last leg only'. If you untick that you can see every flight that any given aircraft has done within approx the previous 24 hours. It's remarkable how much they're worked (although unsurprising given the cost of your average Airbus 320 etc). They often are turned around within a few hours, the engines will be barely cooled.
I wonder whether pilots take on a plane that someone else has just landed and have to deal with their mess or stench in the cockpit/sleeping areas etc.

What thepurist said, the curvature of the earth is measurable over things as small as two hypothetical skyscrapers over the road from each other being further apart at the top than at ground level. But they have to be hypothetical as they'd warp in the sunshine and wobble too much in the wind to actually measure it, but you can put a number to it in the order of millimeters. So it is worth at least thinking about whether it's worth calculating.

The curvature of the earth is about 125mm/km (obviously 2km is more than 250 because it's a curve), but you're only talking 20 ish feet over 28 miles, relative to your "approximately 40,000".

Related fun fact - particle accelerators are built "flat" because you're trying to observe a particle (which should be affected by gravity) that's behaving like a wave (which isn't, not at this scale or frame of reference anyway).

Oh thank gawd. I thought this was going to contain conveyor belt theory.

Same. When WAS that thread? Early days of STW for me, 15+ years ago?

Also, on https://globe.adsbexchange.com/:...

zoom right out and ask yourself how much difference that electric car is going to make.

Assuming it’s at 40,000 feet, how far away from me is the point on the earth it’s directly above?

Distance is as stated. If you want to get there, you may have to travel further.

FYI, there’s a little button on the cog symbol called ‘last leg only’. If you untick that you can see every flight that any given aircraft has done within approx the previous 24 hours

That is awesome. I also spend too much time on there and that button will now mean I spend more.
My current geeky hobby to pass work time is to track military flights heading into the Lake District and have the Windermere webcams open on another screen to try and catch them flying along the lake. Geek overload but it passes the time.

curvature of the earth is 4" in a mile from memory

It was pondering stackexchange that got me wondering. 28 miles away from me is London and Heathrow, so there's a fair amount in the sky. I was wondering if what I was looking at on the map was what I was looking at in the air, given that I really didn't know how high it might be either. just guessing tiny plane = passenger jet at cruising altitude. (or is a smaller plane, closer?) Perspective of flight path is funny at that kind of distance, as well.

Cheers all, and cheers tinas for the curvature maths!

What thepurist said, the curvature of the earth is measurable over things as small as two hypothetical skyscrapers over the road from each other being further apart at the top than at ground level. But they have to be hypothetical as they’d warp in the sunshine and wobble too much in the wind to actually measure it, but you can put a number to it in the order of millimeters. So it is worth at least thinking about whether it’s worth calculating.

I was watching an excellent documentary about the building of the Severn Bridge, and they had to take it into account for the calcs for the towers and road bed, as the tops of the towers, if they were each built 'vertically', would have diverged a bit.

I was wondering if what I was looking at on the map was what I was looking at in the air, given that I really didn’t know how high it might be either

For that I recommend the Augmented Reality mode on the Flightradar24 app - point your phone at the sky and it overlays tags of flight data onto whatever you can (theoretically) see on the screen, then you can see where they are on the map, flightpath etc.

I was watching an excellent documentary about the building of the Severn Bridge, and they had to take it into account for the calcs for the towers and road bed, as the tops of the towers, if they were each built ‘vertically’, would have diverged a bit.

I think they are built vertically aren't they? Meaning the tops are further apart than the bases. Same as the Humber bridge.

EDIT: Here you go. They were moved around during construction to ensure they would be vertical when the deck was finished.

Using da power of siunse I reckon the spot that the plane would vertically over would be slightly closer dude the divergence between vertical for you and verticle for the plane. See my computer model render above

I think they are built vertically aren’t they? Meaning the tops are further apart than the bases. Same as the Humber bridge.

EDIT: Here you go. They were moved around during construction to ensure they would be vertical when the deck was finished.

Yeah, er, that's what I meant. I think 😉

^ no accounting for refraction of light. Poor. 😉

Awesum renderization. U got Silicon Graphicz?

My current geeky hobby to pass work time is to track military flights heading into the Lake District and have the Windermere webcams open on another screen to try and catch them flying along the lake. Geek overload but it passes the time.

Approved!

https://globe.adsbexchange.com/ < Alternatively you can just look up the plane here and get real time data. Its 2021 kids, you don’t need to use your brain anymore.

If only there were some way to look up the date on the internet...

I thought this was going to contain conveyor belt theory.

This is presumably the order that items on the conveyor belt in Aldi are arranged in order to achieve optimal bag filling post-payment, and therefore ease of unpacking once in the domestic kitchen? If the curvature of the earth is the joker in the above calculations, then the joker in conveyor belt theory would be the presence of offspring, or indeed spouse, at any stage of conveyor belt loading or unloading. (Aka as 'WTF are the chocolate digestives? Which moron packed them under 5lb of potatoes?')

Nah bruv, I use solid works, coz it is solid, innit

And presumably, works.

Given the quality of the source data (an assumed height which could be a mile out and inclination derived from "rough hand measurements"), I doubt that curvature of the Earth is a primary factor in this calculation.

We're all mathematically accurate but practically a back door off. "Somewhere between 20 and 30 miles away" is probably at an accuracy still best defined as "hopeful."

It'll be much further that by now

Given the quality of the source data (an assumed height which could be a mile out and inclination derived from “rough hand measurements”), I doubt that curvature of the Earth is a primary factor in this calculation.

We’re all mathematically accurate but practically a back door off. “Somewhere between 20 and 30 miles away” is probably at an accuracy still best defined as “hopeful.”

Thanks! On top of all of all the inaccuracy, I've got a Maths degree with a couple of courses in projective geometry. But I'm a little bit busy today, and I know everyone on here loves planes and maths and knew I'd have more fun reading a thread about my ponderings than doing it on my own.

I'm off to download the flightradar app now. That's the afternoon gone. Good job I got the urgent stuff done this morning!

If only there were some way to look up the date on the internet…

Heh!

Thanks! On top of all of all the inaccuracy

Wasn't meant as a pop at you, rather that folk calculating it down to hundredths of a mile (ie, less than 20 yards) felt a bit ambitious. 😁

Using Flightradar24 I am amazed at how far away planes cruising at 30000ft are still visible - or the progressing front of their vapour trail at least.

I’m in Surrey and can see them heading over The Channel in the south or Luton in the north.

I’m in Surrey and can see them heading over The Channel in the south or Luton in the north.

Previous bit of "what can I see?" I did was trying to work out what white blob I could see on the horizon from the Devils Punchbowl viewpoint in Hindhead. Bit of Google earth sleuthing later revealed it to be a NATS radome at Bovingdon, nearly 42 miles away.

An in retrospect, that should have been a decent clue to the fact that, for the purposes of this exercise, I could safely ignore the curvature of the earth.

I think you are all wrong.

It's 394.5526586km (calculated)

Or it's 393.7990322km (scale drawing)

Edit, that's based on eye level 2m above sea level. And yes it's obviously wrong 😅

I am amazed at how far away planes cruising at 30000ft are still visible

I'm in North Bucks and a couple of summers ago, when the sun was low in the evening, I simultaneously watched and FR tracked a plane at full height (30000+ ft), and as I lost it in the distance (and it got too low in the sky) it was just crossing the coast over South Wales heading to 'Murica. According to the web that's 188 miles straight line distance.

Horrible question but I’ve always wondered…if a plane at full cruising height and speed were to explode how far would the wreckage travel?
Would that plane in the OP (assuming it’s travelling directly towards the observer) travel that far?
If a plane is directly overhead is it safe to think the wreckage wouldn’t come straight down but carry on forwards?

if a plane at full cruising height and speed were to explode how far would the wreckage travel?

Depends on what size the bits of wreckage are. A quick google shows that the Lockerbie wreckage was scattered over 2000km2

[img] [/img]

It's about 44km.

You need to separate the 15 degrees into the angle down to the horizon from horizontal and the angle up to the plane from the horizontal.

Then from the perspective of the viewer you have the line of site line equation and the equation of the circle all based around the origin at the centre of the earth.

You then work out the theoretical intersection points and then you can calculate the angle from person to centre of the earth to the plane.

And with that you convert it to radians and multiply it by the radius of the earth.

I got my angles jumbled in my calculation

The bits start accelerating vertically downwards die to gravity and decelerating horizontally due to air resistance.

However if it exploded stuff could be accelerated backwards effectively increasing the deceleration rate. Potentially it could fall vertically if all the horizontal forces balance instantaneously or light large surface area items that are affected by the wind or can glide in some way. Engines etc will fall in some sort of parabola I imagine.

The Lockerbie disaster if phenomenally unlucky for a whole plane directly hit a rural town when the crash wasn't related to runway approach or takeoff.

According to the web that’s 188 miles straight line distance.

Apparently, I'm on a fight path between USA and Paris. I lose them over the top of the hedge when they're about 5 miles off the coast near Eastbourne, a mere 63 miles away. It is quite a high hedge, though.

I was going to say what you actually need is you eye elevation and the elevation of your horizon.

Was my thinking correct though?

No clue! Sounds good. I'm in a dip, surrounded by trees, so as well as guessing the the altitude of the plane, and the elevation above the horizon, I'm also having to guess where the horizon point is. I do know roughly how high my eyes are, though. Roughly.

My mental maths - and not peeking!- as I'd use it in my routine within aviation:

Using the 1 in 60 rule, If it's 1º above the horizon at 60nm, it's 1nm away vertically, at 15º then 15nm above.

40'000' is approx 6.6nm (1nm = approx 6000'). 6.6nm is just under half of 15, so therefore it's roughly 30nm.

Refining it a little further mentally, 6.6/15 = 0.44. 0.44x60 = 26.4nm. Which if you multiply by 1.15 = 30 statute miles as near as dammit.

That's all pretty rough maths and doesn't account for dipping angles (curvature).

1 in 60 rule super-useful for working out heights of cloud tops using the radar amongst other things like navigation etc..

Flightradar will give you a quicker (and more accurate) answer! 😂