Testwiki:Reference desk/Archives/Mathematics/2017 April 28

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April 28

line of sight from building height

There's a simple formula for calculating your line of sight given a certain height H: $d\approx 3.57\cdot \sqrt{H}$.

But what about the longest possible line of sight given two heights, H₁ and H₂? (Assuming spherical Earth of course) ECS LIVA Z (talk) 07:42, 28 April 2017 (UTC)

Just realized that when H₁ = H₂ it's simply twice of $3.57\cdot \sqrt{H}$. That's pretty cool. ECS LIVA Z (talk) 07:46, 28 April 2017 (UTC)

I think it might be $3.57\cdot \sqrt{H1}+3.57\cdot \sqrt{H2}$ but I'm not sure. ECS LIVA Z (talk) 08:35, 28 April 2017 (UTC)

Template:Ping Two heights of what?--Jasper Deng (talk) 09:30, 28 April 2017 (UTC)

Buildings, spherical cows, or really tall people would work too. ECS LIVA Z (talk) 10:09, 28 April 2017 (UTC)

Yes, that's right, $3.57\cdot \sqrt{H1}+3.57\cdot \sqrt{H2}$. In each case these longest sightlines are tangent to the Earth's surface. The longest sightline between the two heights is tangent to the surface at exactly the same place where the sightlines to the horizon from either of the two heights would be, so it's the same line and its length equals the sum of the horizon distances from the two heights. (Assuming not only spherical Earth, but no refraction of light, of course.) --76.71.6.254 (talk) 10:19, 28 April 2017 (UTC)

Thanks! ECS LIVA Z (talk) 10:31, 28 April 2017 (UTC)

Surely the formula is not dimensionless? —Tamfang (talk) 09:28, 1 May 2017 (UTC)

The given link your line of sight given a certain height H says "the height is given in metres, and distance in kilometres". PrimeHunter (talk) 10:47, 1 May 2017 (UTC)

Indeed. The full formula is:

$d\approx \sqrt{2Rh}$

where R is the radius of the Earth, 6,371 km. If we measure h in metres and we want d in km, then the constant has units of

$\sqrt{10^{6}m}$

and a value of

$\sqrt{2\times 6.371}\approx 3.57$

Gandalf61 (talk) 10:58, 1 May 2017 (UTC)

number of independent sets in a regular graph

Hi all,
How many independent sets of size k are there in a d-regular graph on n vertices?
Thanks in advance — Preceding unsigned comment added by 185.120.126.40 (talk) 18:10, 28 April 2017 (UTC)

It depends on the graph, not just on d, n, k. (Compare the hexagon with two disjoint triangles, for example, with k = 3.) --JBL (talk) 20:06, 28 April 2017 (UTC)

Is there some lower bound on this number? (For the case the graph is connected) — Preceding unsigned comment added by 185.120.126.103 (talk) 08:58, 29 April 2017 (UTC)

Sure: it is certainly at least $n(n-d-1)\cdots (n-(k-1)(d+1))/k!$ -- after you've chosen j points, they and their neighbors make a set of size at most j(d+1) that you can't choose from again. (This is sharp exactly when the graph is a disjoint union of complete graphs.) --JBL (talk) 12:59, 29 April 2017 (UTC)

Space of connected acyclic graphs on N

Consider the space of all connected acyclic graphs whose vertices are the elements of N. A basic clopen set is of the form $[F]$ where $F$ is an acyclic graph on $\{0,1,\dots ,n\}$, and $[F]$ is the set of all connected acyclic graphs that restrict to $F$ on $\{0,1,\dots ,n\}$. Is this space Polish?

The natural metric is $d(G_0,G_1)=2^{-n}$, where $n$ is largest such that the restrictions of $G_0$ and $G_1$ agree on $\{0,1,\dots ,n\}$, but this isn't complete. For example, for $i$ odd, let $H_i$ be the graph with an edge between $x$ and $x+2$ for every $x$, and an edge from $0$ to $i$. Then $d(H_i,H_j)=2^{-\min (i,j)+1}$, but the limit isn't connected.--2406:E006:2C7:1:5CE9:32F9:4BCD:AEF3 (talk) 21:31, 28 April 2017 (UTC)

Answering my own question: yes. Define $d(G_0,G_1)=2^{-n}$, where $n$ is largest such that for every $i,j\le n$, the (unique) path from i to j in $G_0$ is identical to the corresponding path in $G_1$.--2406:E006:2C7:1:10A1:247E:F704:923A (talk) 22:19, 28 April 2017 (UTC)