Entry tags:
Dragon Maths check
I 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?
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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.
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