De-wobbling Tables

Helen sent me this:

I'm not sure where this originated, so apologise for the lack of a reference. I'll happily update if someone can point me in the right direction! In the mean time I'll just link to the tweet I was sent.

I've heard this before: if you're suffering from a wobbly four-legged table don't mess around with trying to fold a napkin or stack beer mats to just the right thickness. Instead just turn the tables on that problem by... turning the table. Twist it around its own axis of symmetry and, within a quarter turn, you'll have a stable table.

There's a reason why this works and one of the places you can find out about that is in the Numberphile video embedded below:

Venus Transits and the Size of the Solar System

This entry in 2020's Big Lock-Down Math-Off hosted by the Aperiodical introducess the Venus transits of the 1600s and their importance in developing an early understanding of the true scale of our Solar system (when the page loads, scroll about half way down to find the article).

Venus Transits and the Size of the Solar System: https://aperiodical.com/2020/05/the-big-lock-down-math-off-match-23/

And you can find out more about Venus at Blogstronomy! 

How Can We Calculate the Angular Size of an Object in the Sky?

This post is written as a sister-post to this one over at Blogstronomy, and is intended to show more explicitly the mathematical process that is alluded to there.

Angular size, as I describe on the Blogstronomy post, is a way that astronomers use to talk about how big things appear in the sky. That post talks about how big Jupiter and its moons Io and Ganymede would appear in the sky above Europa. This post explains how to work it out.

In order to work it out we need to be fairly comfortable with trigonometry - that's that GCSE topic with SOH CAH TOA and right-angled triangles. We also need to know two things about the object we want to look at: its actual size, and its distance from us.

Here's the general maths that we need:

Drawn by me using Paper on the iPad

The grey lines are imaginary lines from our eyeball to the extents of the object in the sky. r stands for the radius of the object we're looking at, and the yellow line (d) represents the distance between us and the centre of the object. Luckily for us, it also cuts the (grey) isosceles triangle in half and gives us a right-angled triangle that we can get to work on with trig.

If we're trying to find the angle that's marked (that's half the angular size), we can label the radius as the 'opposite' side, the yellow line as the 'adjacent' side, and the grey line is the hypotenuse. Let's call the angle "x", to make things easy. We know the radius and distance, so from our school-days "SOH CAH TOA", we know that we have to use the Tangent function:

Tan(x) = Opposite / Adjacent

Or, using more sensible notation from the diagram:

Tan(x) = r / d

We want to find x, though, so we can rearrange:

x = Tan^-1(r / d)

But if we're being honest with ourselves, we want to find the entire angle between the two grey lines, so we need to double whatever result we get (calling the angular size "a"):

a = 2x = 2 x Tan^-1(r / d)

Now all we have to do is substitute the values we know for each of the objects. Here's the data you need:

• Io: r = 1830 km ; d = 249,300 km.
• Ganymede: r = 2631.2 km ; d = 399,300 km.
• Jupiter: r = 69,911 km ; d = 671,100 km.

All you have to do is whack those numbers in the right places into your favourite calculator, then we've found the angular size of these three bodies as viewed from Europa! Just make sure your calculator's set to 'degrees' and neither 'radians' nor 'stun'. The result for Io comes out at 0.841 degrees. If you want to find out what that means for our view and/or find out the answers for the Ganymede and Jupiter (but try them out yourself first, or it's no fun), head over to the post at Blogstronomy!