# World Builder's Handbook Feedback

This is comically down the line from the original release, but...

Someone on the discord asked about the Tidal Heating Factor formula for moons, and if they'd done their computation right. In trying to help them, I suggested plugging in the values for Earth's moon and seeing if they matched what was given in the book as a way to test if they were using the right figures.

From the WBH, Page 126:

The formula given in the book is:
`(M² * S⁵ * e²)/(3000 * a⁵ * P * m)`
Where:
• M is the Primary's Mass (in M⨁)
• S is the World (i.e.: Moon's) size code
• e is the Moon's eccentricity
• a is the Moon's semimajor axis in millions of km
• P is the Moon's orbital period in days
• m is the World (i.e.: Moon's) mass (in M⨁)
For our Moon, those values are: M = 1; S = 2; e = 0.0549; a = 0.384399; P = 27.321661; m = 0.0123
Which when plugged into the equation give us:
`(1² * 2⁵ * 0.0549²)/(3000 * 0.384399⁵ * 27.321661 * 0.0123) = (1 * 32 * 0.00301401)/(3000 * 0.00839288448 * 27.321661 * 0.0123) = 0.09644832/8.46144839481 = 0.01139855914`

Which is a fair bit off of the given value of 52 for Luna.
Assuming that the formula given in the book as-presented is already incorporating the "divide by 3000" step, we still find a value of ≈ 34.195 and not the 52 given in the text.

I must assume I am doing something wrong, though what exactly I know not.
See the post above. I was being too clever by far...

I was wrong about being wrong…

Trying to reconstruct what I did 2+ years ago (with no notes, just spreadsheet scribbles)

Assumption One:

The factors I included for tidal energy change in the following manner: ~M^2*R^5* N* E^2/A^6

M=mass of planet

R = radius of world (or Size in this unitless case)

N = mean angular speed in orbit

E= eccentricity

A = orbital distance

From that, I considered N to be proportional to the circumference of the orbit / orbital period, so that’s proportional to 2*pi*A / P, where A is the orbital distance and P is the period… since the fudge factor to make the temperature work out is going to be a constant added later, we can ignore the 2*pi and we can reduce the power of A in the divisor from 6 to 5.

And this is the total amount of tidal energy.

So, Assumption Two:
The energy is going to be applied to the entire body and the actual temperature increase is therefore going to be proportional to the mass of the body heated – so 1/ Mass = the actual temperature increase on the body from tidal heating. (If you follow my assumption, all that energy goes into the object and it is applied in direct proportion to the mass – this is my weakest assumption, but there it is - I have a hindsight sneaky suspicion that it might be more correct to assume 1/ Mass^(1/4) ) .

So this gives the final formula as written:

~M^2*R^5* E^2/(A^5 * P * Mass of moon)

For Luna the values I plugged in were:

M =1

R = 2.17175 (here is where you’re going to get a difference for sure, since I took the radius of the moon divided by 1600 times 2 rather than just 2.)

E = 0.055 (not sure why I wasn’t so precise here, but at least I wasn’t off a full decimal point)

A= 0.3844

P= 27.322

M = 0.0123

Which should get you 51.81315 which rounds to 52, and then divide by 3000 to get 0.017271.

Is this correct? Don’t know. A real astrophysicist would probably complain a lot about my assumptions (especially the second one), but if you look at the Wikipedia article https://en.wikipedia.org/wiki/Tidal_heating (not my only source, but I never found anything more useful) it talks about things like “the imaginary portion of the second-order Love number)” and nope, I just wanted to get a number that gave a plausible temperature that varied in a plausible way based on values we already had.
Heh, no worries - less concerned about a PhD thesis and more concerned about consistency and reproducibility, and this helps greatly.

Now that I know that you were using diameters to compute size, rather than just use size, the pieces fall in to place, lol.

So to sum up, the disconnect seems to have been a few typos (including the orbital distance), and ambiguity on the actual values used. Perfect, I feel less crazy now

Thanks very much - I really appreciate you taking the time to dust off the cobwebs

Here is a little mnemonic for remembering the sequence of spectral type letters:
Oh Be A Fine Girl Kiss Me

Here is a little mnemonic for remembering the sequence of spectral type letters:
Oh Be A Fine Girl Kiss Me
Or, if you are from a slightly different background, Oh Boy An F Grade Kills Me...

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