Why is precession angular velocity of a gyroscope independent of the angle of its axis?

I'm making gyroscopes in my shop and studying precession. Studied dynamics in 1964 so may have forgotten a thing or two. Consider a gyroscope or top spinning with its axis at angle $$\beta$$ with vertical. The (correct?) equation for angular velocity of precession is: $$\Omega = mgd/I\omega$$ where "r" is linear distance to center of mass. Apparently independent of angle β, which is angle of the axis from vertical. Seems to me it should be $$\Omega = mgrsin\beta/I\omega$$. Otherwise $$\Omega$$ would be no different when axis is vertical ($$\beta=0$$) compare with axis horizontal. What am I missing here? In subsequent post, I can attach a photo of my gyroscope.

Thanks

EDIT: Let's replace "d" with "r". Attached is drawing. Also shown is description which shows equations. Still do not understand how precession angular velocity can be independent of angle theta.

I'm not sure I understand the description that you give for what 'd' represents.

It could be that in the source that you use that factor 'd' is used for the horizontal component of the distance between the pivot point of the gyroscope, and the center of mass. The larger that value, the larger the torque.

I recommend you supply the following diagram:

Draw a gyroscope and indicate with two arrows where you place the start point and the end point of the length factor that you refer to as 'd'

(It's probably helpful to supply two diagrams, the first one showing the gyroscope aligned with the vertical, the second one showing the gyroscope at an angle to the vertical. That may well help to prevent any ambiguity.)

Other than that:
There is my 2012 answer where I discuss how gyroscopic precession works. The discussion that I present there does not use the concept of angular momentum vector, nor does it use vector operations such as vector cross product. Instead the discussion capitalizes on symmetry.