Dragon Maths check
Sep. 17th, 2007 07:35 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
ffuturesI need someone to check a little maths for me.
First read this passage from Jo Walton's web site, describing Dragon units of time:
http://www.zorinth.net/bluejo/books/dragon/dates.htm
Assuming that the dragon foot is the same as our own, and that gravity is the same – both open to question – it’s possible to convert the figures on that page to determine the length of the day and year compared to our world.
If I've done my maths right, the dragon second is 1.864 human seconds, the time it takes a drop of water to fall 30ft. from rest at 1g. The other Dragon time units convert as follows:
1 dragon minute = 80 dragon seconds
= 149.1 earth seconds
1 dragon hour = 80 dragon minutes
= 198.83 earth minutes
1 dragon day = 20 dragon hours
= 66.27 earth hours
= 2.76 earth days
1 dragon year = 200 dragon days
= 1.51 earth years
Okay... if I've got all that right, and assuming that this world has a climate similar to our own, it must be in a wider orbit than the Earth. Can anyone give me figures for the orbital radius, and if possible a ballpark estimate of what type of star it would need to be?
First read this passage from Jo Walton's web site, describing Dragon units of time:
http://www.zorinth.net/bluejo/books/dragon/dates.htm
Assuming that the dragon foot is the same as our own, and that gravity is the same – both open to question – it’s possible to convert the figures on that page to determine the length of the day and year compared to our world.
If I've done my maths right, the dragon second is 1.864 human seconds, the time it takes a drop of water to fall 30ft. from rest at 1g. The other Dragon time units convert as follows:
1 dragon minute = 80 dragon seconds
= 149.1 earth seconds
1 dragon hour = 80 dragon minutes
= 198.83 earth minutes
1 dragon day = 20 dragon hours
= 66.27 earth hours
= 2.76 earth days
1 dragon year = 200 dragon days
= 1.51 earth years
Okay... if I've got all that right, and assuming that this world has a climate similar to our own, it must be in a wider orbit than the Earth. Can anyone give me figures for the orbital radius, and if possible a ballpark estimate of what type of star it would need to be?
![[identity profile]](https://www.dreamwidth.org/img/silk/identity/openid.png){kind=small}
no subject
Date: 2007-09-17 07:26 pm (UTC)Depends on the type of star. If the star is the same as ours then the orbit will be 1.31AU.
(See http://en.wikipedia.org/wiki/Orbital_period)
However, the much longer day is likely to rule out the same climate as there will be, for starters, more opportunity for night time cooling and thus greater differences between day and night time temperatures leading to all sorts effects.
no subject
Date: 2007-09-17 07:31 pm (UTC)no subject
Date: 2007-09-17 07:37 pm (UTC)no subject
Date: 2007-09-17 08:02 pm (UTC)Then assume both seconds and and feet are the same and work out gravity from that.
Now pick the answer you like the look of.
no subject
Date: 2007-09-17 07:56 pm (UTC)no subject
Date: 2007-09-17 07:28 pm (UTC)no subject
Date: 2007-09-17 07:36 pm (UTC)no subject
Date: 2007-09-17 08:00 pm (UTC)However, another solution is to simply assume the dragonworld's sun is a sun-like G2-star but older, because then of course it would be brighter. But then it probably is so old it is moving off the main sequence into a subgiant, G2IV-V, and then the planet would be almost 10 billion years old.
You can use a middle way, say an F9V star older than the Sun but just an extra billion years or so, if my math works out.
no subject
Date: 2007-09-17 09:19 pm (UTC)no subject
Date: 2007-09-17 10:27 pm (UTC)no subject
Date: 2007-09-17 11:03 pm (UTC)g is 981cm sec squared, so after a second the object is travelling at 981cm per second.
To reach that speeed it has to start from zero and accelerate, so the average speed in the first second is
981 - 0 / 2, only 490.5 cm/sec
After two seconds the average speed is 981cm cm/sec, and by then it has travelled more than 30ft, so the the correct time for a 30ft fall is a bit less than 2 seconds.
no subject
Date: 2007-09-17 11:05 pm (UTC)no subject
Date: 2007-09-18 11:06 am (UTC)no subject
Date: 2007-09-18 11:53 am (UTC)no subject
Date: 2007-09-18 04:47 am (UTC)no subject
Date: 2007-09-18 05:59 am (UTC)