I was looking at the cloud cover and wondered how high up it is. Is there some mathematical formula you can apply, perhaps based on something on the ground, that can be a general guide to estimate height of clouds, or planes, or whatever?
In short, no.
With clouds you can have a good guess based on the type of clouds. Different types of clouds occur in certain altitude bands and if you know the type of cloud you can have an educated guess at a range of heights for the cloud. Or, if you live in a hilly area and the cloud is below the tops of the hills you could know the height based on knowledge of the height of the hills.
Gavin Pretor-Pinney of the Cloud Appreciation Society has published a good book called The Cloud Spotter’s Guide.
With aeroplanes you could work it out based on apparent size and the aircraft type, but it’s not something you could do on the spur of the moment and I’m not sure how accurate you could be.
Put a vertical stake in the ground, right below some easily recognisable feature of the object of which you want to measure the height. Put a second stake in the ground, some distance away from the first one, aimed at the same feature. Measure the distance between the two stakes and the angle of the second stake.
Then, use basic trigonometry: the tangent of an angle in a rectangular triangle is equal to the opposite side divided by the adjacent side. So you multiply the tangent of the angle by the distance between the two stakes, and you have your answer.
The greater the distance between the two stakes, the better the accuracy of the answer, until you start placing them so far away that you need to start taking the curvature of the earth into account.
In practice, this can be difficult with clouds (because they don’t have small, sharply defined features) or airplanes (because they move around while you’re setting up the stakes). But the principle works.
Turns out that “rectangular triangle” is a bit too literal a translation from my native language. The English term is “right triangle” or “right-angled” triangle. It’s a triangle in which one of the angles is 90 degrees.
But don’t you also need to know the horizontal distance to the cloud, or whatever? Knowing the angle is no use if you don’t know the base of the triangle.
The base of the triangle would be the line between the two stakes.
Yeah, Walton Firm was presuming that the first stake was directly below the cloud (or whatever it is you’re measuring.) The method would work in principle, but it does seem really impractical unless you’re measuring a small solid object that’s hovering in the sky.
I need a funky laser doodad that can reflect off clouds.
A laser ceilometer? Not exactly cheap.
And pilots really dislike having lasers pointed in their vicinity. It’s an offence to do so here in the U.K.
I said that not realising there was such a thing. So never fear, I will not be shining lights in pilots’ eyes.
Yeah, I missed that part. I thought he meant using two stakes some distance away and then scaling up using similar triangles.
Cloud base can be calculated with:
Cloud Base Altitude = (((temperature - dew point) / 4.5) * 1000) + measure station altitude, assuming you have that info handy. Any other object is largely guesswork.
They start at about $4k, IIRC.
In 1963, the local military where I was would make a pilot go up and see each morning. Worked good.
Most aviation reporting places now have stuff to get it pretty close although some observers can do the low stuff really good with just looking.
Also Pilot reports in real time help tune things.
Okay, if and only if it’s an item whose size you know (say, a Boeing 737-800, which is what WestJet flies —39.5m), it shouldn’t be too difficult to get a rough guess. This assumes, of course, that you can view the plane from exactly its side.
You need a meter-stick with a clear metric ruler mounted perpendicular on the end, such that you hold the meter-stick up to your eye and can look at the plane through the clear ruler. As you do so, measure the width of the plane on the ruler, which we’ll call W. Now, W/1m=39.5m/D, where 1m is the distance from your eye to the ruler and D is the actual distance to the plane. Hence, D = 39.5m•1m/W. So you can find the distance to the plane.
Now, somewhere on the meter-stick, you want a pendulum mounted on a protractor — or whatever device you want to measure the angle above the ground that the plane is at. (The protractor would need its 0º-line to be perpendicular to the meter-stick, i.e. perpendicular to the ground when horizontal). Hence, you can take the angle (call it M) of your view of the plane from horizontal. Now, this creates a right triangle — with D being the hypotenuse, horizontal distance to the plane being the adjacent side, and the altitude of the plane being the opposite.
Hence, the plane’s altitude = DsinM = (39.5/W)sinM
Obviously, this requires a very contrived situation (though, again, it’s plausible to identify the plane by knowing that certain airlines only fly certain types), but does anybody see a reason that it wouldn’t work?
Yeah we sometimes get asked for a report on the cloud base so the met observers can confirm their estimates.
The problem with using apparent size as detailed by straight man is that most people don’t can’t tell a 737 from a 767 except that a 767 is bigger so you can’t tell what type it is to begin with without knowing how far away it is, and you don’t know how far away it is without knowing what type it is. Local knowledge of routes and typical aircraft types can help as can knowledge of typical altitudes flown in the area you’re looking at. For example I know that airliners normally lose about 320 feet for every mile traveled when on approach so if my house is 15 nm from the airport and it’s on final approach I know that any airliner flying over will be at about 4800 feet (this isn’t perfect, in many places airliners will fly a short level segment at say 3000 feet until they intercept the ILS glideslope.)