I’ll give it a try:

So from the given data we can write an equation relating current I to speed W:

I = (Istall-Ifree)(1-W/Wfree)+Ifree

We can also write an equation that relates T to W:

T = (Ts)(1-W/Wfree)

Kt = T/I, however if you divide these two equations you get Kt as a function of W, and this is because the equations have different roots. Theoretically they have the same roots as at T=0->I=0, but they do not because the equation for T only takes into account external loads on the motor and ignores the torques due to friction forces in the motor.

So the equation for T really is:

T = (Ts)(1-W/Wfree)-Tloss

Solving I=0 for W gets:

W(I=0) = Wfree(1+Ifree/(Istall-Ifree))

Then plugging that W into Tloss = (Ts)(1-W/Wfree) to solve Tloss gives:

Tloss = -(Tstall*Ifree/(Istall-Tfree))

Plugging Tloss back into T = (Ts)(1-W/Wfree)-Tloss gives us:

T = Ts(1-W/Wfree+Ifree/(Istall-Ifree))

Now since these two equations have the same root, we can divide them and get a constant:

Kt = T(W)/I(W) = -Tstall(Istall/(Ifree-Istall))/Istall

Plugging in for a bag motor yields:

Kt = 0.0102

——————

Ke is much simpler to solve for:

Ke = V/W

However you need to take into account the voltage drop through the motor’s windings at W, so we first need to solve for internal motor resistance which is:

R = Vspec/Istall

We now can rewrite Ke to include this drop:

Ke = (Vspec-I(W)*R)/W

Since we are given Wfree and Ifree we can plug those in along with R and get:

Ke = (Vspec-Ifree*Vspec/Istall)/Wfree

Plugging in for a bag motor yields:

Ke = 0.0078

As to why they are different… I would guess bad measurement, but its more likely I did the calculation wrong