An aircraft built for a planet with 1.38 atmospheres would do just fine. Big storms correlate with high atmospheric pressure, but that's from Earth's perspective- a world with 1.38 atm might have an "excessive" pressure of 145 kPa.
Alternate Planets, Suns, Stars, and Solar Systems Thread
- Thread starter Pragmatic Progressive
- Start date
Hapsburg
Banned
Very well, then. The weather is now the reverse from what was previously stated.
I dunno, I just looked up what was the maximum atmospheric pressure for air flight, and got a lot of stuff talking about temporary flight restrictions being placed when air pressure is above a certain limit. Though yeah I'm sure they can engineer aircraft that are able to operate perfectly fine well in excess of that limit, I was just using that as a baseline.
I was intrigued and had a look, I can find quite a few references to commercial aircraft using barometric altimeters, devices that work out their altitude based on external air pressure and some very generalised assumptions. It might just be related to that - the higher the current atmospheric pressure at sea level above the nominal 1013 mbar, the lower the reading from such an altimeter relative to actual height. The barometer will likely show they're flying lower than they actually are. @Ramp-Rat might know for certain, he has excellent commercial aircraft knowledge.
Ramp-Rat
Monthly Donor
Right it’s been a very long time since I did my airmanship course, and being a work right now I am relying on my memory. With barometric altimeters, it depends on what sort of flying you were doing, locals or distance. If you were flying strictly local, you set the altimeter to zero when on the runway and left it there throughout your flight. If doing distance work, you set the altimeter to zero on the runway, then at a set hight tuned it to a set international mark, thus insuring that all aircraft are showing themselves to be the same hight. That is if the outside air pressure is X at a nominal Y hight, then every aircraft irrespective of its true hight, will be showing that its flying at Y. And according to the clock rules all aircraft flying at Y, will be heading in the same direction.
There are a number of problems, the hight shown is not the true hight above the ground, but a nominal hight above sea level. And with atmospherics, the five thousand feet of clearance between you and the mountain below, might only be two thousand, or less. Running into a rock filled cloud, can really ruin your day. And when you get to the end of your flight you have to get from the ground the local zero altitude barometric setting, especially if you are going to let down through clouds. You feel a right fool if your altimeter is showing that you are five hundred feet up, when in fact you are fifty feet below the ground.
RR.
There are a number of problems, the hight shown is not the true hight above the ground, but a nominal hight above sea level. And with atmospherics, the five thousand feet of clearance between you and the mountain below, might only be two thousand, or less. Running into a rock filled cloud, can really ruin your day. And when you get to the end of your flight you have to get from the ground the local zero altitude barometric setting, especially if you are going to let down through clouds. You feel a right fool if your altimeter is showing that you are five hundred feet up, when in fact you are fifty feet below the ground.
RR.
Deleted member 105545
At 1000 AU, none of the planet's will be greatly effected, though Neptune may experience some minor disruptions. The Kuiper belt will suffer from disruptions as well. The Oort cloud will be greatly disrupted however. Expect comets streaming in toward the inner Planets. Also, Sedna is going to be either shot out of the Solar system, gain orbit around the new star, or shot toward the inner planets.
I don't think disrupted is the right word. The star isn't being put in place today, it's forming in concert with the sun 4.6 billion years ago. Won't the shell of the Oort cloud encompass both stars.
If it's forming in concert, the planet's as we know them are probably butterflied away.
Some of you may well be acquainted with the Planet Map Generator that can be found at the other end of the link posted below - if you have not, then I highly recommend it to one and all, especially to those who love to feed their imaginations on Worlds as yet unnamed and awaiting the words that make somewhere out of something - but even those of you who are not will doubtless be able to make more out of it than I can.
https://topps.diku.dk/torbenm/maps.msp
Quite frankly I have been frustrated for some time now by an inability to work out the precise scale of the various physical features on the maps produced by this excellent device - the better to appreciate how large the various continents & islands & other elements of this planetscape are compared to their Earthly counterparts; may I please ask if someone could help me to work out how to read scale into any Planet I care to generate with this device?
Thank You in advance for your consideration!
https://topps.diku.dk/torbenm/maps.msp
Quite frankly I have been frustrated for some time now by an inability to work out the precise scale of the various physical features on the maps produced by this excellent device - the better to appreciate how large the various continents & islands & other elements of this planetscape are compared to their Earthly counterparts; may I please ask if someone could help me to work out how to read scale into any Planet I care to generate with this device?
Thank You in advance for your consideration!
One thing you can do is generate an equirectangular map, download it, and open it up as a texture on Google Earth. If you use a map of the whole world, it will cover the entire globe, but the resolution might be bad. You can use this as a template if you want to import small, more detailed sections.
Thank You very kindly for the suggestion and for taking the time to help Falkanner - I'm a little intimidated by the fact you need to read FOUR documents before using Google Earth but will serious consider doing so; having said that I'm not so much interested in blowing up sections of the generated maps (it's easy enough to do that in the Planet Map Generator itself) and more interested in learning how to read them more fluently.
Basically I have no real idea of what calculations might be required to work out the size of a planet after establishing a rough scale (in other words if I only have the radius, how does one use that to work out the size of Continent & Oceans et al?); the problem is that I know so little about mapping one doesn't even know the right questions to ask!
Basically I have no real idea of what calculations might be required to work out the size of a planet after establishing a rough scale (in other words if I only have the radius, how does one use that to work out the size of Continent & Oceans et al?); the problem is that I know so little about mapping one doesn't even know the right questions to ask!
... Basically I have no real idea of what calculations might be required to work out the size of a planet after establishing a rough scale (in other words if I only have the radius, how does one use that to work out the size of Continent & Oceans et al?); the problem is that I know so little about mapping one doesn't even know the right questions to ask!
As a quick indicator to help get a feel, it helped me to find a map of the Earth in the same projection as the Planet Map Generator - I left it with the default Mollweide projection. I already knew that the Earth's surface is roughly 70% water but having Earth and the generated map side-by-side gave me a much better feel for the size of the continents, assuming the planet's radius is similar to Earth's.
I don't know of any firm rules for a sensible dry land percentage but as we know Earth 'works' I aim for a similar feel in the ocean to land ratio. That boils down to 'if it's not immediately visible that it's mainly water then I've got too much land'. Then to gauge specific continents I overlay the Mollweide Earth in a half-transparent layer over the generated map and drag it around to park our continents over the map's ones, to get a feel for relative size.
Not that you couldn't have a world with much more land, especially if it's a colonised world as opposed to one where humanoid life evolved (most terraformed Mars maps for example), and if your lifeforms are far from humanoid then using Earth as a guide is far less helpful. But it does help give me a feel for the distances involved
CannedTech
Banned
IIRC he’s cool with it, he gave me permission and a few people used his Europa map
Anyone know if a worlda style has been made of the 'World Beyond the Poles'
Considering that this map is one from the flat Earth files, I'm not sure worlda makes sense for it.
To clarify on the title, K-type main-sequence stars--shortened as "orange dwarves"--are subjects of excitement for astronomers and astrobiologists.
Why?
In this alternate scenario, Earth is the third planet in a binary star system. It still has the moon we know, and it still tilts at an angle of 23.5 degrees. One star is a K-type main sequence star--or "orange dwarf"--that has 80% the mass of a G-type main sequence star--or "yellow dwarf". The first to orbit this star is another orange dwarf, one that has 45% of our sun's mass. In regards to their relationship with Earth, questions follow:
Why?
- They emit enough radiation to provide a high-enough temperature to make water liquid but not high enough for solar radiation. (This would mean no auroras, but it also means no genes would be damaged by UV exposure.)
- They have a longer lifespan than G-type stars. (15-30 billion years compared to our sun's total of ten billion.) This means that life would be given more time to evolve.
- They are three to four times more common than G-type stars, making the search for Earth-like exoplanets hypothetically easier.
- Despite having 45-80% of a G-type star's mass, they can still be just as bright.
In this alternate scenario, Earth is the third planet in a binary star system. It still has the moon we know, and it still tilts at an angle of 23.5 degrees. One star is a K-type main sequence star--or "orange dwarf"--that has 80% the mass of a G-type main sequence star--or "yellow dwarf". The first to orbit this star is another orange dwarf, one that has 45% of our sun's mass. In regards to their relationship with Earth, questions follow:
- How far would the second star orbit the first star to keep the solar system stable?
- In this scenario, Earth is still the third planet, but this time, it's the fourth body to orbit the first star. How close must it orbit both stars to retain its liquid surface water and average atmospheric temperature of 14 degrees Celsius?
- In the equator, how would the stars be positioned at dawn, noon and dusk?
- In relation to #3, what would dawn, noon and dusk look like in the spring, summer, autumn and winter months of the middle and high latitudes?
1. The second star doesn't orbit the first. Because any two stars in the main sequence will be relatively close in mass, both stars will orbit around a common centre of gravity, an imaginary point called a barycentre. Technically, all pairs or bodies, including the Earth and the Sun, are in mutual orbit around the barycentre for that system, but as the Sun is so much more massive than the Earth, this barycentre is still inside the Sun. It's not at the Sun's exact centre, but nor is it far away.
You have two main options here, which informs the answer. Do you want a Close Binary pair, where the stars are close, between 0.15 and 6 AU say, and if you want Earthlike habitable planets then the lower the better within this band. If you're not familiar with the term, AU means Astronomical Unit and is the average distance from the Earth to the Sun, roughly 93,000,000 miles or 150,000,000 km. Then all the planets orbit the stars' common barycentre, basically. This gives you a Tatooine-type sky. The other option is a Distant Binary pair, with a separation of say 120 AU upwards, and 200 - 300 is better for Earthlike worlds. Planets will then orbit each star separately, so star A has a set of planets and star B has its own set.
Based on the rest of the question I assume a Close Binary pair, so the answer is 0.15 to 6 AU, any value between 0.15 and 0.4 should suffice. This puts the barycentre at an average distance (technically semimajor axis distance) of 0.07 AU from star A and 0.13 AU from star B.
2. The answer to this depends on the stars' luminosities and the orbital eccentricities about the barycentre. I've used formulae from a Youtuber called Artifexian who's done a good series of videos on this, restating equations into relative-to-our-Sun type units. The rule of thumb for luminosity is mass cubed, giving us 0.51 of the Sun's for star A and 0.09 for star B. This means the habitable zone, where water will be liquid on a planet very similar to Earth, is closer to the stars than for our Sun.
The guide for eccentricites, how far skewed from a perfect circle the orbits are, is 0.4 to 0.7. 0 is a pefect circle. I picked 0.4 and 0.41 for A and B. The closest they ever come to each other is therefore 0.12 AU, and the furthest apart is 0.28 AU. The pair of stars create a nominal zone, within which their gravity will destabilise a planet's orbit and either tear it apart or fling it out of the system. The inner and outer edges of this zone are 1/3 x Minimum sep. and 3 x Maximum sep. respectively. So no planets could orbit between 0.04 and 0.84 AU. This makes it less likely that you could have a habitable Earth as the third body within this system, but let's plough on.
The base line for the habitable zone is square root of the two luminosities added together. 0.51 + 0.09 is 0.6, √0.6 ≈ 0.7746 AU. The inner and outer edges, using a conservative take on a habitable zone, are 0.95 and 1.37 times the base, so 0.736 to 1.061 AU wide. That will fit Earth at our 1 AU nicely, although we'll be further out, hence colder. We can't move Earth to the baseline itself as that's still inside the riptide zone, which extends out to 0.84 AU. Say we moved it to 0.91, that would still be closer, cooler than we are here but still in the zone, and with a 'year' of 0.776 years or roughly 284 days.
But please remember I've used someone else's simplifications aimed at building systems similar to ours, not necessairly at modelling K-type stars, I'm not an astrophysicist. My mate who is will be dealing with small children right now, it's too early for her to check my working! Sorry for any oversights. If you read up on Ks and find the luminosity to be too low especially then that throws all the habitable zone numbers out as they're all driven exclusively by that.
Questions 3 and 4 I can't answer, sorry. I think you'd be best hunting for one of the star system modelling programs out there as the fancier ones can not only give you better figures than I can, but can also model what things would look like from arbitrary points on the planet's surface. I can't recommend any right now, I've only got a tablet to hand for now so can't install anything decent, it doesn't have the power or the input tools (keyboard, mouse). But the answer will vary depending on where the stars are in their orbit around the barycentre relative to each other and the planet you're on.
You have two main options here, which informs the answer. Do you want a Close Binary pair, where the stars are close, between 0.15 and 6 AU say, and if you want Earthlike habitable planets then the lower the better within this band. If you're not familiar with the term, AU means Astronomical Unit and is the average distance from the Earth to the Sun, roughly 93,000,000 miles or 150,000,000 km. Then all the planets orbit the stars' common barycentre, basically. This gives you a Tatooine-type sky. The other option is a Distant Binary pair, with a separation of say 120 AU upwards, and 200 - 300 is better for Earthlike worlds. Planets will then orbit each star separately, so star A has a set of planets and star B has its own set.
Based on the rest of the question I assume a Close Binary pair, so the answer is 0.15 to 6 AU, any value between 0.15 and 0.4 should suffice. This puts the barycentre at an average distance (technically semimajor axis distance) of 0.07 AU from star A and 0.13 AU from star B.
2. The answer to this depends on the stars' luminosities and the orbital eccentricities about the barycentre. I've used formulae from a Youtuber called Artifexian who's done a good series of videos on this, restating equations into relative-to-our-Sun type units. The rule of thumb for luminosity is mass cubed, giving us 0.51 of the Sun's for star A and 0.09 for star B. This means the habitable zone, where water will be liquid on a planet very similar to Earth, is closer to the stars than for our Sun.
The guide for eccentricites, how far skewed from a perfect circle the orbits are, is 0.4 to 0.7. 0 is a pefect circle. I picked 0.4 and 0.41 for A and B. The closest they ever come to each other is therefore 0.12 AU, and the furthest apart is 0.28 AU. The pair of stars create a nominal zone, within which their gravity will destabilise a planet's orbit and either tear it apart or fling it out of the system. The inner and outer edges of this zone are 1/3 x Minimum sep. and 3 x Maximum sep. respectively. So no planets could orbit between 0.04 and 0.84 AU. This makes it less likely that you could have a habitable Earth as the third body within this system, but let's plough on.
The base line for the habitable zone is square root of the two luminosities added together. 0.51 + 0.09 is 0.6, √0.6 ≈ 0.7746 AU. The inner and outer edges, using a conservative take on a habitable zone, are 0.95 and 1.37 times the base, so 0.736 to 1.061 AU wide. That will fit Earth at our 1 AU nicely, although we'll be further out, hence colder. We can't move Earth to the baseline itself as that's still inside the riptide zone, which extends out to 0.84 AU. Say we moved it to 0.91, that would still be closer, cooler than we are here but still in the zone, and with a 'year' of 0.776 years or roughly 284 days.
But please remember I've used someone else's simplifications aimed at building systems similar to ours, not necessairly at modelling K-type stars, I'm not an astrophysicist. My mate who is will be dealing with small children right now, it's too early for her to check my working! Sorry for any oversights. If you read up on Ks and find the luminosity to be too low especially then that throws all the habitable zone numbers out as they're all driven exclusively by that.
Questions 3 and 4 I can't answer, sorry. I think you'd be best hunting for one of the star system modelling programs out there as the fancier ones can not only give you better figures than I can, but can also model what things would look like from arbitrary points on the planet's surface. I can't recommend any right now, I've only got a tablet to hand for now so can't install anything decent, it doesn't have the power or the input tools (keyboard, mouse). But the answer will vary depending on where the stars are in their orbit around the barycentre relative to each other and the planet you're on.