tag:blogger.com,1999:blog-7028787815719483796.post5931404734861777353..comments2018-08-27T07:20:51.387-05:00Comments on This Orb: Celestial Architect Spreadsheetj. w. bjerkhttp://www.blogger.com/profile/06800512284198234202noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-7028787815719483796.post-20484490854575687882017-10-22T15:14:36.941-05:002017-10-22T15:14:36.941-05:00Fantastic resource! I know it's several years ...Fantastic resource! I know it's several years later, but before I do the conversion myself, has anyone moved this to Google Sheets?Chris Brinkleyhttps://www.blogger.com/profile/14979973221495902944noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-63742036609242866342010-11-04T11:03:13.348-05:002010-11-04T11:03:13.348-05:00I had originally planned to make the spreadsheet c...I had originally planned to make the spreadsheet calculate the albedo based on user input, but ran into some problems.<br /><br />If you would like to collaborate on this chart as a co-author, contact me by email (accessable in my blogger profile)j. w. bjerkhttps://www.blogger.com/profile/06800512284198234202noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-25305151953699693522010-11-04T09:50:10.795-05:002010-11-04T09:50:10.795-05:00Well, you need to start somewhere to get a ballpar...Well, you need to start somewhere to get a ballpark figure for a planet’s temperature. Insolation, albedo, and a stab at the greenhouse factor are the simplest model you can do. A more detailed model will give the user control over the fraction of the albedo due to oceans, forest, ice, fields, deserts and overall cloud cover. It sounds daunting but it’s really very doable in a spreadsheet. You can see, for example, how the albedo goes down as the oceans get bigger- water absorbs most of the light hitting it. Bare rock, on the other hand, may reflect 20% or more of the light back into space. You multiply the fraction of the planetary surface that is terrain type X by the albedo for that terrain type, and then sum all the terrain types to arrive at the total albedo.<br /><br />Technically the greenhouse factor belongs under the radical as well (as a scale factor for the effective wattage rather than the effective temperature- note that you must take the fourth power of the greenhouse effect if you move it under the radical with the other energy terms), but you can get away with it as a legitimate fudge factor as written above if you are careful and understand what it means. For Earth, the greenhouse factor is 1.1294- the ratio between the observed mean surface temperature 288 K and the radiative equilibrium temperature of 255 K. In other words, it’s an empirical observation about how much various climate feedbacks raise the temperature by trapping outgoing IR radiation and reflecting it back to the surface. <br /><br />On Earth, all these factors combine to raise the temperature by 13% (or to raise the effective surface flux by 62.7%, if you prefer). One presumes that human-habitable planets will have greenhouse factors that range from about 10% to 15%- outside of this range you get into atmospheric pressures too low to breathe on one end, and carbon dioxide levels too high to breathe on the other. I might push those boundaries to 5%-20% under special conditions, but no further than that.<br /><br />Runaway greenhouses will raise the effective surface temperature by a factor of e, the base of natural logarithms (2.7182818…), but no more than this. See Venus for an example.Orionhttps://www.blogger.com/profile/08818431929827830349noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-77983985032007730292010-11-01T14:05:02.661-05:002010-11-01T14:05:02.661-05:00Thanks for the clear error report Orion. I'll...Thanks for the clear error report Orion. I'll try to fix that. I don't remember exactly which source(s) i used, i think i ended up adapting something on the edge of my understanding.<br /><br />My math is weak enough that the only way i catch equation errors is by comparing the result to a known planet-- but temperature on a real planet is so complex that the numbers wouldn't match anyway.j. w. bjerkhttps://www.blogger.com/profile/06800512284198234202noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-30793507778333325772010-10-30T05:47:32.188-05:002010-10-30T05:47:32.188-05:00I'm concerned about the temperature calculatio...I'm concerned about the temperature calculation you have implemented. For a blackbody, the basic equation is T = 65 * (W^0.25), where T is in Kelvins, W is expressed in watts per square meter, and 65 is just a scaling constant to end up in units of kelvins. Planets, of course, are a little more complicated than ideal blackbodies, but they're not impossible to tackle. <br /><br />In cell D48, you take the fourth root of the insolation before subtracting the albedo correction. The order of mathematical operations is incorrect. The albedo correction must be applied to get the net wattage before taking the fourth root. The constant 374 doesn't make much sense- this is clearly just the wattage at which a blackbody radiates into space at 288 K (i.e., Earth's average temperature), and it does not belong in the equation. You need to go back to first principles to understand why this shortcut won't work.<br /><br /> The full equation should look like:<br /><br />T = 65 * (1+G) * ( 340*L/(R^2) * (1-A) )^0.25<br /><br />where G is the greenhouse increment, L is the star's luminosity relative to Sol, R is the planet's semi-major axis in AU, A is the planetary albedo, and 340 is the solar constant averaged over the surface of a rotating sphere (i.e., 1360 W/sq m divided by 4, since spheres have four times the surface area of the circle of starlight that they're intercepting in space). The calculated temperature will be in kelvins.<br /><br />The main point is that a body's temperature changes as the fourth root of the effective rate at which the body receives energy; all energy terms in the equation must fall under the radical. It should be clear from this why trying to use Earth's effective surface wattage as a fudge factor to scale the temperature of another planet with a different effective surface wattage cannot give you the correct mathematical answer.Orionhttps://www.blogger.com/profile/08818431929827830349noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-87521851558017032412010-05-28T17:00:49.666-05:002010-05-28T17:00:49.666-05:00A superb program.
*downloading a copy to use* tha...A superb program.<br />*downloading a copy to use* thank you for making and sharing this.<br /><br />have nice days and be well.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-72060565169548816882010-04-24T16:58:38.955-05:002010-04-24T16:58:38.955-05:00Metaraptor:
No, you should do each moon individual...Metaraptor:<br />No, you should do each moon individually. You should be able to duplicate that page if you want-- or just write the info down somewhere else, and then put in the info for your next moon.<br /><br />Each moon should be at a different distance from the planet.<br /><br />Multiple moons would be more or less unaffected by each other. Technically they would tug on each other a little, but the difference would be negligible.<br /><br />Multiple moons would have a cumulative effect on the tides-- in rather complicated ways. In short when moons are lined up, or on opposite sides of the planet their pull would create especially high tides.j. w. bjerkhttps://www.blogger.com/profile/06800512284198234202noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-70479558635458163992010-04-22T21:32:10.219-05:002010-04-22T21:32:10.219-05:00Just another question regarding the Celestial Arch...Just another question regarding the Celestial Architect Spreadsheet. The moon and planet section seems to be based around a planet with a single moon. What if one wants to do a planet with more than one moon? Would they just add the combined mass of the moons?Metalraptorhttps://www.blogger.com/profile/17053007518293924808noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-76091806863500541882010-04-14T21:48:16.796-05:002010-04-14T21:48:16.796-05:00Metalraptor:
Technically no. The average temperat...Metalraptor:<br />Technically no. The average temperature is based on 4 other variables (since this is a simplification). So an average temperature of 20º could be achieve by different combinations of closeness to the sun, larger or smaller suns, and different levels of albedo and a greenhouse effect. There's not only one combination that will yield 20º.<br /><br />But practically, if you choose values for 3 of the variable, you could switch around the formula to do that, or simply raise or lower the last variable until the temperature is what you want.j. w. bjerkhttps://www.blogger.com/profile/06800512284198234202noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-1422278186168281142010-04-14T21:24:28.058-05:002010-04-14T21:24:28.058-05:00Just wondering, would it be possible to "reve...Just wondering, would it be possible to "reverse-engineer" the values in the Celestial Architect Spreadsheet so you could enter the average temperature of the planet, and work backwards from there?<br /><br />(Sorry, spelling errors)Metalraptorhttps://www.blogger.com/profile/17053007518293924808noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-53775065816573155392010-04-07T09:00:29.372-05:002010-04-07T09:00:29.372-05:00Thanks, i'm glad you find it useful.
You misu...Thanks, i'm glad you find it useful.<br /><br />You misunderstand exactly what the 'Habitable Zone' means. It's a ring around the star where a habitable planet might exist. You can have a habitable zone without any planets in it. I updated the text of this post to make that clearer.<br /><br />You are right that mass and atmosphere are important in determining whether a planet is actually habitable. Mass and density are dealt with in this post: http://orb.jwbjerk.com/2009/11/how-heavy-how-big.html<br /><br />I'd love to provide more info about how mass effects the atmosphere, but i simply haven't been able to find it.j. w. bjerkhttps://www.blogger.com/profile/06800512284198234202noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-36672300602418121472010-04-07T03:49:07.980-05:002010-04-07T03:49:07.980-05:00oops, sorry, i just noticed that you do have a pla...oops, sorry, i just noticed that you do have a planet size/density calculator, its just tucked away in another tab. this is my first time using excel in ages. but even so, it does not seem that the habitable zone beginning and end is affected by this change in mass.Michaelhttps://www.blogger.com/profile/01872249986874708144noreply@blogger.comtag:blogger.com,1999:blog-7028787815719483796.post-89108853754975288762010-04-07T03:41:19.199-05:002010-04-07T03:41:19.199-05:00this is an amazing little spreadsheet, but it omit...this is an amazing little spreadsheet, but it omits a major part of the determination of a habitable zone: the planets mass, (which in order to make things realistic and simple, should be derived from its size and overall composition) and atmosphere. the reason for this is the definition of the habitable zone.<br /><br />the habitable zone, as im sure you know, is where liquid water can exist and stay as liquid water (some would argue that life could exist outside the habitable zone, just going dormant during summer or winter, whenever water goes out of its desirable phase). gravitational and atmospheric pressures make up a large part of where the boiling point lies, and thus where the habitable zone lies.<br /><br />for example, a higher gravity planet would presumably attract a more dense atmosphere, creating very high pressures. these combined pressures would make it harder for water to boil, resulting in a higher boiling point, and thus a planet which requires to be closer to its star than the average earth sized planet.<br /><br />i am currently working with a planet thats roughly mars sized (probably larger, im terrible with these calculations, so im not sure if it would be realistic), thus having lower gravity and a thinner atmosphere. because of this it would probably be further out from its star, at the higher end of what this spreadsheet determines is my habitable zone. im not sure how large or massive this orb is, so im not sure if you have to worry about it that much.Michaelhttps://www.blogger.com/profile/01872249986874708144noreply@blogger.com