I used to be able to do these

@venda
To be unambiguous: Can the “straight line” be planet-planet-sun OR planet-sun-planet?.

When will it next occur? Ah ok assuming same planet planet sun....


Ok I think I found a way that will be labor intensive for 6,5 and 2. In 1 year each planet will travel 5/30, 6/30 and 15/30 of its orbit. This is like ring theory where every complete time around the ring starts all over again. So you just keep adding a fraction for each year throwing put any whole number that results. Of course this would only find a whole year answer easily.

@iamatiger said
@venda
To be unambiguous: Can the “straight line” be planet-planet-sun OR planet-sun-planet?.

Yes, either

@venda said
At one time I could do this type of thing easily.
Now apparently not because recently I came across this and I can't work out how to get to the answer so I must be going wrong somewhere.The given answer doesn't agree with what I get it to.
I'm sure someone here can solve the puzzle and show me the working out please:-
A man walks from one town to another.
On the first day h ...[text shortened]... urth day half of the remaining distance.He now has 14 miles left.
How far has he travelled?

Just a point on this type of problem. They are deliberately constructed to be confusing. They tend to require fluid intelligence, which is more of a young persons thing, as one ages crystallized intelligence (roughly speaking experience) becomes more important. In so far as they are useful it is because they force one to be careful in analysing data, which in real world problems often isn't that neat and tidy.

@deepthought said
Just a point on this type of problem. They are deliberately constructed to be confusing. They tend to require fluid intelligence, which is more of a young persons thing, as one ages crystallized intelligence (roughly speaking experience) becomes more important. In so far as they are useful it is because they force one to be careful in analysing data, which in real world problems often isn't that neat and tidy.

I agree.I used to teach my children to look out for traps and also to think about what the answer should be approximately so as to avoid quoting an answer where the working out may be at fault.
A good example is simple multiplication.
I learned my times tables at school but they're not taught in the modern day.
My daughter would rely competely on her calculator, so if for instance she acidentally inputted a wrong number she would not know that she had the wrong answer

OK, I think I have a solution.

I found a formula for planetary alignment they way I was assuming (all planets on a single radius away from the sun):

For orbits of period p1, p2, p3, where p1 < p2 < p3:
alignment of p1 with p2:
t = k / (1/p1 - 1/p2), where k is any integer

alignment of p1 with p3:
t = j / (1/p1 - 1/p3), where j is any integer

The three planets are aligned with both equations are satified:
(1/p1 - 1/p3) / (1/p1 - 1/p2) = j / k

Substituting our values for p1, p2, and p3
(1/2 - 1/6) / (1/2 - 1/5) = j/k
j/k = 10 / 9

back into one of the original equations, let's use j
t = 10 / (1/2 - 1/6)
t = 30

That's the number I also got and showed earlier. This problem states that alignment doesn't need to be in a radius, however.

I think we can solve this by halving the orbital periods. If the planets are orbiting twice as fast, then instead of being across from each other they would be aligned instead. Solving again:

j/k is still 10/9

t = 10 / (1/1 - 1/3) = 15

I think this means that the solution is always half of the radial alignment solution? It doesn't seem quite right to me, but I can't see why not.

@forkedknight said
OK, I think I have a solution.

I found a formula for planetary alignment they way I was assuming (all planets on a single radius away from the sun):

For orbits of period p1, p2, p3, where p1 < p2 < p3:
alignment of p1 with p2:
t = k / (1/p1 - 1/p2), where k is any integer

alignment of p1 with p3:
t = j / (1/p1 - 1/p3), where j is any integer

The three plan ...[text shortened]... half of the radial alignment solution? It doesn't seem quite right to me, but I can't see why not.

Well done.
Thank you
15 was the answer given.
I'll now look at the other 3 planet ones in the book and use your equations.
I'm reasonably sure you're correct

To simplify:
@venda I think your equation for two planets is correct, you just have to solve for j/k instead of applying the equation again for a third planet.

t = j * p1 * p3 / 2 (p3 - p1)
j / k = (p2*p3 - p2*p1) / (p2*p3 - p1*p3)

The blog/page I found with the planetary equations also has a formula for syzygy, which is planetary alignment, not necessarily including the sun / center of orbit:
https://www.mathpages.com/home/kmath161/kmath161.htm

I thought it would be interesting to plot your example problem:
https://imgur.com/tgF8lLj

The planets are aligned when the plot crosses 0

Not sure how well these links will work on this forum, but I thought it was interesting.

@forkedknight said
That's not quite true.

I was talking about the same alignment occuring after X years,
where X is the lowest common multiple of the two orbital times.

@wolfgang59 said
I was talking about the same alignment occuring after X years,
where X is the lowest common multiple of the two orbital times.

Yes, I got what you meant, but @venda stated in the initial prompt:
> They do not have to be in the same position as at the start to be in a straight line.

*edit*
My point was that alignment along a single radius does not require a LCM. The question about alignment along a radius or along a diameter (out of phase 180*) is somewhat separate.

Two planets that are lined up PPSun will generally line up PSunP in half the time it takes them to get back to PPSun.

If the orbital period of the slow planet is S and the fast planet period is F then the next PSunP will happen when the fast planet has done exactly half an orbit more than the slow planet.

T/F = T/S + 1/2
1/F-1/S = 1/(2T)
T = FS/(2S-2F)

So this should be soluble with the same techniques

@iamatiger said
Two planets that are lined up PPSun will generally line up PSunP in half the time it takes them to get back to PPSun.

If the orbital period of the slow planet is S and the fast planet period is F then the next PSunP will happen when the fast planet has done exactly half an orbit more than the slow planet.

T/F = T/S + 1/2
1/F-1/S = 1/(2T)


So this should be soluble with the same techniques

T = FS/(2S-2F)

What this is saying is that the planets will be in line with the sun whenever
T=JFS/(2S-2F)
Where J is an integer

Using ForkedKnight’s cunning approach we can add a third planet M: M and F will line up when
T = FS/(2S-2F)

What this is saying is that the planets will be in line with the sun whenever
T=KFM/(2M-2F)

and therefore all 3 planets are in line when
KFM/(2M-2F) = JFS/(2S-2F)

Remove common factor of F/2 and put in terms of J/K
J/K = M(S-F)/(S(M-F)
=(M-MF/S)/(M-F)
=(1-F/S)/(1-F/M)
So with periods of 1, 3 and 6 we get line ups when
J/K = (5/6)/(2/3) = 15/12 = 5/4

{Notice that in reducing the fraction to its simplest form we have done something equivalent to finding the highest common divisor}

So line ups occur at J/K=N*5/4 where N is any integer

i.e when K is a multiple of 4 and J is a multiple of 5

Since
T=KFM/(2M-2F)
We have
T={0,4,8,12,16} * 3/4 = {0,3,6,9,12...}

Trying these out
At T0 the planets are lined up SMFX where X is the sun
At T3, F=3, M=1, S=1/2: line up is MFXS
At T6, F=6, M=2, S=1: line up is SMFX
At T9, F=9, M=3, S=3/2: line up is MFXS
At T12, F=12, M=4, S=2: line up is SMFX

@iamatiger said
T = FS/(2S-2F)

What this is saying is that the planets will be in line with the sun whenever
T=JFS/(2S-2F)
Where J is an integer

Using ForkedKnight’s cunning approach we can add a third planet M: M and F will line up when
T = FS/(2S-2F)

What this is saying is that the planets will be in line with the sun whenever
T=KFM/(2M-2F)

and therefore all 3 planets are ...[text shortened]... 1: line up is SMFX
At T9, F=9, M=3, S=3/2: line up is MFXS
At T12, F=12, M=4, S=2: line up is SMFX

Wow this is all deep stuff Tiger.
As I said I would do, I got the old book out again and after going thro' Forkedknights equations and satisfying myself that they worked I found another example with the planets orbiting in 1,7 and 49 years.
I replicated Forkedknights method but didn't get the right answer!.
However , interestingly when I looked at the answer , I had written in the margin all those years ago " this is wrong" the answer is .....
Working backwards from the answers(mine and the one given) I found that both answers were a solution to the alignment(I can't remember how I came up with the alternative answer allthose years ago).
I also looked at the SYZGY site mentioned by forkedknight.
Waaay too complicated for me I'm afraid