Created
February 12, 2015 19:46
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Calculate Kerbin-Duna Aldrin cycler periods in KSP
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def calculate_periods(precision=1e-1, periods_to_check=1000, print_first_n=7): | |
"""Calculate optimal orbital periods for a Kerbin-Duna cycler. | |
This assumes that the cyclers will have the same orbital period as the | |
synodic period. Multiples of this period are also acceptable, but the | |
results will need to be modified accordingly. | |
All arguments are optional. | |
Args: | |
precision: Acceptable absolute error. Default is 0.1 (10.65 days) | |
periods_to_check: Number of future cycles to check. Default is 1000. | |
print_first_n: Detail the first n successful cycles. | |
Returns: | |
A list of the cycles in which the cycler returns to Kerbin. | |
Examples: | |
>>> periods = calculate_periods(print_first_n=2) | |
7 cycles (14.9421 years): Kerbin would be -58.83 SOI radii behind. | |
8 cycles (17.0766 years): Kerbin would be 77.81 SOI radii ahead. | |
""" | |
kerbin_year = 9203545 # seconds | |
duna_year = 17315400 # seconds | |
synod = (1/(1/kerbin_year - 1/duna_year)) / kerbin_year # ~2.135 years | |
kerbin_vel = 9284.5 # m/s around Kerbol | |
kerbin_soi = 84159286 # m | |
years_count = synod*np.arange(periods_to_check) | |
offset = years_count - np.round(years_count) # How far are we off from a full year? | |
good_cycles = np.where(np.isclose(offset, 0, atol=precision)) # See where it is close. | |
# Print the results | |
for results in good_cycles: | |
for i, n_cycles in enumerate(results): | |
if i == 0: # Skip the initial result (all zeros) | |
continue | |
if i > print_first_n: # Stop printing | |
break | |
n_years = synod*n_cycles | |
dist = (n_years - np.round(n_years)) * kerbin_year * kerbin_vel / kerbin_soi | |
direction = "ahead" if dist > 0 else "behind" | |
time_string = "{:d} cycles ({:.4f} years):".format(n_cycles, n_years) | |
dist_string = "Kerbin would be {:.2f} SOI radii {}.".format(dist,direction) | |
print(time_string, dist_string) | |
return good_cycles |
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