  Steve Jackson Games Forums [SPACE] Tidal braking
 Register FAQ Calendar Mark Forums Read  08-05-2006, 08:08 AM #1 Agemegos   Join Date: May 2005 Location: Oz [SPACE] Tidal braking erratum On p. 117 of Space, in Step 30 of the planet generation sequence, we estimate the tidal distortion of a planet as T = (0.46 * M * D) / R^3, (where M is the mass of the star, D is the diameter of the world, and R is the radius of the world's orbit). And the total tidal effect (amount of slowing) is E = (T * A)/ m (where A is the age of the system and m is the mass of the world). That expression expands to E = (0.46 * M * D * A) / (R^3 * m). Now suppose that the system is only just old enough that the planet has tide-locked. Substituting E = 50 (the critical value for tide-lock), we get (0.46 * M * D * A) / (R^3 * m) = 50, so A(lock) = (109 * R^3 * m) / (M * D) That's the time to tide-lock in GURPS Space Look at the formula for time to tidelock given in Wikipedia. This is t(lock) = (w * R^6 * I * Q)/(3 * G * M^2 * k * (D/2)^5) (making the appropriate substitutions for the different notation.). I (moment of inertia of a spherical planet) is approximately = 0.1 m D^2. G, Q, and k are constants, the latter two having to do with the material of the planet. w is the initial rotation rate of the planet. Collect the constants t(lock) = (0.1 * Q)/(3 * G * k / 32) * (w * R^6 * D^2 * m)/(M^2 * D^5) = constant * (w * R^ 6 * m)/(D^3*M^2) Compare A(lock) = (109 * R^3 * m) / (M * D) Something is obviously wrong. I suspect that the authors of Space overlooked the following: that the tide height is proportional to teh tide-raise force divided by the surface gravity; that the tidal torque is proportional not only to the height of the bulges, but to the gravity gradient (0.5*G*M/r^3) and the moment arm (D times the sin of the angle between the bulges and the star-planet line); and that the tendency of the planet to resist rotation, its moment of inertia, is proportion to its mass times the square of its radius. If I (and Wikipedia) are right, we ought ot replace the current expression for E with E = (0.46 * A * D^3 * M^2) / (R^6 * m) Which should make it a lot easier to get planets aroung type K stars that are not tide-locked. Which in turn should be greatly to Zorg's relief. __________________ © copyright Brett EvillFLAT BLACK texts on-lineDiscussion of FLAT BLACK Last edited by Agemegos; 10-30-2010 at 08:33 PM. Reason: compacted the layout to improve readability   08-05-2006, 11:22 PM #2 sir_pudding Wielder of Smart Pants   Join Date: Aug 2004 Location: Ventura CA Re: [SPACE] Tidal braking You should PM (or email) Jon F. Zeigler.   08-12-2006, 01:15 AM #3 Agemegos   Join Date: May 2005 Location: Oz Re: [SPACE] Tidal braking Here is a website that gives a comprehensible derivation for the equilibrium height of the tidal bulge. Its result is h = (3 * M * r^4)/(m*R^3), where M is the mass of the tide-inducing body, r is the radius of the world, m is the mass of the world, and R is the distance to the tide-raising body. For simplicity, we want this in the form h = k * (M * D^4)/(m*R^3) To find k in terms of the relative units in Space we substitute the values for the Earth-Moon system, noting that lunar tides produce about 70% of the observed tidal range: 0.7 = k * (0.0123 * 1^4)/(1*30.1^3) k = 1.55 million. A moon with mass M times Earth's mass circling a planet with mass P times Earth's mass and diameter D times Earth's diameter at distance R times Earth's diameter will raise a tide T times as high as the tides on Earth, where T = 1.55 million * (M * D-to-the-fourth) / (P * R-cubed) To find k in terms of the other set of relative units used in Space we subsitute the values in the Earth-Sun system, noting that solar tides make up about 30% of the tidal range on Earth: 0.3 = k * (1 * 1^4)/(1 * 1^3) k = 0.3 A planet with a mass P times Earth's mass and D times Earth's diameter orbiting a star of mass M solar masses at a distance of R astronomical units will experience a solar tide with a range T times as high as the total tidal range on Earth, where T = 0.3 * (M * D-to-the-fourth) / (P * R-cubed) Unless anyone has something to add I'll submit a corrigendum to errata@sjgames.com in a couple of days. __________________ © copyright Brett EvillFLAT BLACK texts on-lineDiscussion of FLAT BLACK Last edited by Agemegos; 10-13-2007 at 11:47 PM.   08-12-2006, 02:33 AM #4 Agemegos   Join Date: May 2005 Location: Oz Re: [SPACE] Tidal braking I have been carefully through the derivation and got a result that is consistent with the result in Wikipedia. The mass of a tidal bulge is proportional to the area of the bulge (D squared) times its height (h), times the density of the distorted body (P). That comes out to mass of tidal bulge = P * h/D The tidal acceleration is GMr/R^3, or since we are ignoring constants tidal acceleration = MD/R^3 F = ma, so Tidal braking force = h * M * P / R^3 Torque is force times moment arm, and for a given angular displacement of the tidal bulge the moment arm is proportional to D torque = h * M * P * D / R^3 The moment of inertia of a sphere is 0.4 * D^2 * m, so ignoring constants I = D^2 * P angular acceleration = torque/I = (h * M) / (R^3 * D) substituting h = (M * D^4)/(P * R^3) gives angular deceleration = (M^2 * D^3) / (P * R^6) E = A times angular acceleration E = (k * A * D^3 * M^2) / (R^6 * P) Woot! I love it when a plan comes together! Now there remains to chose the most convenient form for the equation and estimate the constant of proportionality. Given that the Ms and Rs are in different units according to whether we are calculating lunar or solar tides, but the D and P are always the same, the most convenient form might turn out to be angular deceleration = h^2 * P/D^5 Which gives E = (k * A * m/D^5) * sum of squares of tide heights. Now to estimate k. __________________ © copyright Brett EvillFLAT BLACK texts on-lineDiscussion of FLAT BLACK Last edited by Agemegos; 08-12-2006 at 02:37 AM.   08-12-2006, 06:05 AM   #5
Agemegos

Join Date: May 2005
Location: Oz Re: [SPACE] Tidal braking

Quote:
 Originally Posted by Agemegos angular deceleration = h^2 * P/D^5 Which gives E = (k * A * m/D^5) * sum of squares of tide heights. Now to estimate k.
Let's assume that Earth started out with a period of 12 hours, which is pretty typical for freely-rotating bodies in the Solar System. That's a rotational speed of 2 revolutions per day. Earth's period is now 24 hours and its rotational speed 1 per day. E in Earth's case is 1 rotation per day.

1 = (k * 4.5 * 1/1^5) * (0.7^2 + 0.3^2)

k = 0.383 revolutions per day.

So, the reduction in spin rate

E = (0.383 * A * P/D^5) * sum of squares of tide heights

in REVOLUTIONS PER DAY lost from the original rotation speed of, say 3d/5 revolutions per day.
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Discussion of FLAT BLACK

Last edited by Agemegos; 10-13-2007 at 11:59 PM.   08-14-2006, 05:36 AM   #6
zorg
Experimental Subject

Join Date: Jan 2005
Location: saarbrücken, germany Re: [SPACE] Tidal braking erratum

Quote:
 Originally Posted by Agemegos If I (and Wikipedia) are right, we ought ot replace the current expression for E with E = (0.46 * A * D^3 * M^2) / (R^6 * m) Which should make it a lot easier to get planets aroung type K stars that are not tide-locked. Which in turn should be greatly to Zorg's relief.
Hm. I seem to have deleted most of my original attempts since they didn't work out as expected, so I couldn't run them through your modified formula (computers...). But I used your formula on the notes I have kept, with confusing (for me) results (note that I have Incomptence: Mathematics, so I may have made stoopid errors).

A small Rock Planet on an inner orbital radius which got a Tidal Effect of 467,82 with the system in Space came out with a Tidal Effect of 1,37482470000e-12 using your formula. I have no idea whether this is better, since I am unable to make sense of this number...

Could you perhaps post the changed formulas together, in an easy to understand post, so that even a number-impaired guy like me can use them? (iow, without your reasoning - just the formulas and a note what each variable is supposed to be, if it differs from Space p.117).
I'm afraid I got something mixed up here... Before I tackle the remaining notes I have here, I would like to be sure that I understood/used your formulas correctly.

I feel a little slow :|
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Last edited by zorg; 08-14-2006 at 06:17 AM.   08-14-2006, 07:44 AM   #7
Agemegos

Join Date: May 2005
Location: Oz Re: [SPACE] Tidal braking

Quote:
 Originally Posted by Zorg Could you perhaps post the changed formulas together, in an easy to understand post
When calculating the height of the tide that a planet raises on its moon, or that a moon raises on its planet:

T = 1.56 million * (M * D-to-the-fourth) / (P * R-cubed)

Where M is the mass of the object raising the tide (in Earth masses), D is the diameter of the object the tides are on (in Earth diameters). P is the mass of the object the tides are on (in Earth masses), and R is the distance between them (in Earth diameters).

T is the range of the tides in terms of the total tidal range on Earth, ie. if T is 1, tides will be about the same as on Earth; if T is 2 the tides will be about twice the size of the tides on Earth.

When calculating the height of the tides raised by a star on a planet or the moon of a planet:

T = 0.3 * (M * D-to-the-fourth) / (P * R-cubed)

Where M is the mass of the star in solar masses, D is the diameter of the planet (or moon) in Earth diameters, P is the mass of the planet (or moon), and R is the distance between them in AU.

T is the range of the tides in terms of the total tidal range on Earth, ie. if T is 1, tides will be about the same as on Earth; if T is 2 the tides will be about twice the size of the tides on Earth.

In the case of an object under the tidal influence of more than one neighbour, add the T values together for a total (maximum) tidal range (ie. spring tides). But keep a record of the separate T values, because you will need them in the next step.

To determine the rotation rate of a planet or moon, roll 3d and divide by 5. The result is the initial rate of rotation of the body (in revolutions per Earth day). Call this value I.

Now square each of the T values to which this world is subject, and add all of the squares together. Call this value S. Then calculate the minimum rate of rotation, w, as follows.

W = I - [(0.536 * A * P/D^5) * S]

Now calculate the maximum daylength

max_Daylength =1/W,

in Earth days. Compare this to the world's orbital period around the body that raises the highest tide on it, or it's orbital period around the body that raises teh highest tide on it. If the orbital period is shorter or the rotational period is negative, then the world is tide-locked to that body. Set the rotational period equal to the orbital period.
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Discussion of FLAT BLACK

Last edited by Agemegos; 10-13-2007 at 11:53 PM. Reason: updating figures   12-28-2006, 05:06 PM #8 Agemegos   Join Date: May 2005 Location: Oz Re: [SPACE] Tidal braking It looks as though no-one cares to get this right. I suggested it as an erratum four months ago, and nothing has happened. Last edited by Agemegos; 11-14-2007 at 12:20 AM.   12-28-2006, 05:09 PM #9 Kaell   Join Date: Aug 2004 Re: [SPACE] Tidal braking I appreciate the effort put into this, and if I ever get a copy of Space I'll be using your erratum wether official or not.   12-29-2006, 05:48 PM #10 balzacq   Join Date: Nov 2005 Location: Seattle, Washington Re: [SPACE] Tidal braking For what it's worth, I put your formula into my PHP star system generator... __________________ -- Bryan Lovely My idea of US foreign policy is three-fold: If you have nice stuff, wed like to buy it. If you have money, wed like to sell you our stuff. If you mess with us, we kill you.   Tags planets, space, system generation, tidal braking, tide, tide-locked, world generation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Fnords are Off [IMG] code is Off HTML code is Off Forum Rules
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