Snails on a Plane Example Calculation

# Example for https://puzzling.stackexchange.com/questions/126002/the-alien-snails-experiment



from math import *



def get_log10_str( log10_val ):
    intpow = floor(log10_val)
    frac = 10**( log10_val - intpow )
    while frac < 1.0:
        frac *= 10
        intpow -= 1
    return f"{frac:.4}∙10^{intpow}"



anne_pos0 = ( 0, 0 )
bill_pos0 = ( 1e12*cos(radians(-24)), 1e12*sin(radians(-24)) )

anne_dir = ( 1, 0 )
bill_dir = ( cos(radians(71)), sin(radians(71)) )

print("Starting configuration:")
print(f"  Anne pos=⟨ {anne_pos0[0]}, {anne_pos0[1]} ⟩")
print(f"       dir=⟨ {anne_dir[0]}, {anne_dir[1]} ⟩")
print(f"  Bill pos=⟨ {bill_pos0[0]:.3e}, {bill_pos0[1]:.3e} ⟩")
print(f"       dir=⟨ {bill_dir[0]:.4}, {bill_dir[1]:.4} ⟩")
print(f"  Speed (μm/s): {1e6*1/86400:.4}")
print()



#Speed described by 1/t, where t the remaining time, so displacement by
#   \int_x^86400 dt/t = ln(86400)-ln(x) = ln(86400/x)
#For higher precision, if x=c*10^p, then
#                     = ln(86400) - ln(c) + p*ln(10)
#                     = ln(86400/c) + p*ln(10)

print("Displacement with time remaining:")
print(f"  t=       60: {log(86400.0/    60.0):.4}")
print(f"  t=       10: {log(86400.0/    10.0):.4}")
print(f"  t=        1: {log(86400.0/     1.0):.4}")
print(f"  t=     1e-3: {log(86400.0/  1.0e-3):.4}")
print(f"  t=   1e-100: {log(86400.0/1.0e-100):.4}")
print(f"  t=1e-100000: {round( log(86400.0/1.0)+100000*log(10) )}")
print()



#Bill crosses x-axis.  The elapsed time is
#   \int_x^86400 dt/t = ln(86400)-ln(x) = `param`

param = -bill_pos0[1] / bill_dir[1]
cross_pos = ( bill_pos0[0]+param*bill_dir[0], bill_pos0[1]+param*bill_dir[1] )

ln_tc = log(86400) - param
log10_tc = log10(e) * ln_tc

print("Bill crosses the x-axis at:")
print(f"  Location      : ⟨ {cross_pos[0]:.3e}, {cross_pos[1]} ⟩")
print(f"  Remaining time: {get_log10_str(log10_tc)}")
print()



#Anne finds trail on x-axis and changes speed to 2/t.  This happens when Anne's displacement is the
#crossing point, so:
#   (cross_pos[0]) = \int_x^86400 dt/t   =>   ln(x) = ln(86400) - (cross_pos[0])

ln_t1 = log(86400) - cross_pos[0]
log_t1 = log10(e) * ln_t1

dist = cross_pos[0]
anne_pos1 = ( dist, 0 )
bill_pos1 = ( bill_pos0[0]+dist*bill_dir[0], bill_pos0[1]+dist*bill_dir[1] )

log_v = log10(2) - ln_t1*log10(e)

separation = hypot( anne_pos1[0]-bill_pos1[0], anne_pos1[1]-bill_pos1[1] )

print("Anne finds the trail:")
print(f"  Anne's position : ⟨ {anne_pos1[0]:.3e}, {anne_pos1[1]} ⟩")
print(f"  Bill's position : ⟨ {bill_pos1[0]:.3e}, {bill_pos1[1]:.3e} ⟩")
print(f"  Separation      : {separation:.3e}")
print(f"  Remaining time  : {get_log10_str(log_t1)}")
print(f"  Anne's new speed: {get_log10_str(log_v)}") #the effect of the 2 is not visible at this scale
print()



#The relative speed of Anne and Bill is -1/t.  They intersect when that speed removes the distance
#between them.  That is:
#   \int_x^exp(`ln_t`) -dt/t = -`separation`

ln_t2 = ln_t1 - separation
log_t2 = log10(e) * ln_t2

log_relvel_by_c = log10(e)*( separation - ln_t1 - log(299792458) )

param = log(86400) - log_t2
pos2 = ( bill_pos0[0]+param*bill_dir[0], bill_pos0[1]+param*bill_dir[1] )

print("Meetup:")
print(f"  Position      : ⟨ {pos2[0]:.3e}, {pos2[1]:.3e} ⟩")
print(f"  Remaining time: {get_log10_str(log_t2)}")
print(f"  Relative speed: {get_log10_str(log_relvel_by_c)} ⨯ c")
print()