SomeConversionFormulae

Here I gather some conversion formulae and other geenral use formulae.

1) Apparent magnitude, absolute magnitude and size.

This conversion depends on 2 main parameters, the distance from object to Sun
and observer, and the phase angle.

r:     Sun-object distance in AU
delta: Observer-object distance in AU

without phase effect, the conversion would be:

m_0 = 5*Log(r*delta) + H

Here, m_0 stands for magntiude at 0 phase angle.

robs:  Sun-observer distance in AU

The phase angle alpha is defined as:

alpha = acos((r^2+delta^2-robs^2)/(2*r*delta))

There are several ways to account for the phase effect. The one I use mostly is
the Bowell's H-G formalism:

m = m_0 - 2.5*Log((1. - G)*phi1 + G*phi2)

with G the slope parameter, originally thought to be in the range 0 to 1,
typically ~0.15 for asteroids, but seems for be negative for many TNOs, ~-0.12,

phi1 = exp(-3.33*(tan(alpha/2.0))^0.63)

phi2 = exp(-1.87*(tan(alpha/2.0))^1.22)

One can also use the linear phase effect as described in Schaefer & Rabinowitz
(2002) and Sheppard & Jewitt (2002):

m = m_0 + 2.5*beta*alpha


More generally, the apparent magnitude is determined from geometrical factors
and the intensity of light source, i.e. of the Sun:

m_X = m_Sun_X - 2.5*Log(p_X*R^2* Phi(alpha)/(2.25*10^16*r^2*delta^2))

where m_X is the apparent magnitude in filter X, m_Sun_X the apparent magnitude
of the Sun in filter X, p_X the albedo in filter X, R the radius of the object
in km, r and delta the Sun-object and observer objects in AU. Phi(alpa) is the
phase angle effect, with Phi(0) = 1.

For Bowell's H-G formalism, Phi(alpha) = (1. - G)*phi1 + G*phi2, while for
linear effect, Phi(alpha) = 10^(-beta*alpha).

Using the previous notations, we get:

m = m_0 - 2.5*Log(Phi(alpha)),

m_0 = m_Sun - 2.5*Log(p*R^2/(2.25*10^16*r^2*delta^2))
    = m_Sun - 2.5*Log(p*R^2/(2.25*10^16)) + 5*Log(r*delta),

hence

H = m_Sun - 2.5*Log(p*R^2/(2.25*10^16))

The usual value of m_Sun is -27.1 in filter R.


2) Rate of motion on sky.

One could in principle write down a general equation for this, but it would be
an intrinsic equation which would require solving Kepler's equation. In
practice, the easiest way to compute the rate of motion on sky is to perform a
numerical derivative, i.e. compute the positions at 2 slightly different times
(2 hours difference for example), and then divide the position difference by
the time difference.

In case of an object on a circular orbit in the plane of motion of the Earth,
and assuming the Earth is on a circular orbit, one can compute the apparent
motion on sky.

a_K:     Semimajor-axis (radius) of KBO orbit (AU)
a_E:     Semimajor-axis (radius) of Earth orbit (AU)
omega_K: Angular mean motion of KBO
omega_E: Angular mean motion of Earth
theta_S: Earth-Sun-KBO angle

d theta_a / d t = (omega_K - omega_E)*(a_K^2 - a_K*a_E*cos(theta_S))/(a_K^2 +
a_E^2 - 2*a_K*a_E*cos(theta_S)) + omega_E

where d theta_a / d t is the apparent rate of motion of the KBO with respect to
the stars.


3) Single, double and divot exponential distribution functions.

3.1) Single exponential.

d N(H) = A*10^(alpha*H) and H between H_0 and H_1. Let Xi be a uniform random
number between 0 and 1,

H = 1/alpha*Log(Xi*(10^(alpha*H_1) - 10^(alpha*h_0)) + 10^(alpha*H_0))

3.2) Double exponential.

N(<H_0) = 0
H_0 <= H < H_1,  d N(H) = A*10^(alpha_1*H)
H_1 <= H <= H_2, d N(H) = B*10^(alpha_2*H)

For a double exponential, A*10^(alpha_1*H_1) = B*10^(alpha_2*H_1), while for a
divot, A*10^(alpha_1*H_1) = C*B*10^(alpha_2*H_1).

... more later ...