## (as in Struik's Notes to NAPIER.)

Let TS = A=107.
Let dS = y.
The position of g at time t is A-y from its starting point, I will let P=A-y.
The position of the token a at time t is bc and is called x.
Let the initial velocity of g be denoted v0.
Then  dP/dt = - dy/dt.
Napier's statement is that " a geometrically moving point approaching a fixed one  has its velocities proportionate to its distances from the fixed one."

Thus at time t, dy/dt = k y where k is a constant for proportionality. Napier is using  k= -1.
Thus y =  v0 e-t . Solving this equation for t using the natural logarithm gives t=ln(v0/y)= ln(v0)-ln(y)

Now since the token a is moving at a constant rate, namely v0, we have x = v0*t.  x is Napier's log for the y value at the same time t. Hence Napier's log for y, which I will denote by NOG(y) has  NOG(y)=v0*(ln(v0/y)) = v0*(ln(v0)-ln(y)).
Notice that for y = A, NOG(y) = 0,  so v0=A.

Notice that the computation using NOG is more complicated then those using the modern logarithms (when A is not 1).

Multiplication Note:

Find the product of two numbers x and y.

Find  NOG(xy) and then use the table to find xy. This is the same as how ln would be used. However the formula used to compute NOG(xy) is not as simple as ln(xy)= ln(x) + ln(y).

NOG (xy) = A* (ln(A) - ln(x) - ln(y))
= NOG(x) -A*ln(y)
= NOG(x) + NOG(y) - A*ln(A).

Using A = 10000000,   ln(A) is approximately 16.118095651

Notice that if a/b=c/d then

NOG(a)- NOG(b)=v0*(ln(b)-ln(a)) = v0*(ln(d)-ln(c)=NOG(c)-NOG(d)

thus
NOG(c)= NOG(a) + NOG(d) - NOG(b)