Boost logo

Geometry :

Subject: Re: [geometry] Rtree in a Hierarchical Method
From: zaz1588 (zaz1588_at_[hidden])
Date: 2014-02-20 17:05:33

Adam Wulkiewicz wrote
> By hierarchical method do you mean that you'd like to traverse the rtree
> internal structure, i.e. internal nodes and leafs by yourself?
> If the answer to the above is yes, then there of course is a method. But
> it's not a part of the 'official' interface and is hidden in the
> details. Since it's in the details, it may be changed in the future.
> That's why it's prefered to use the official interface whenever
> possible. Maybe if you shared what you wanted to do it could be possible
> to find different solution?

My application is completely different but the classical example for the
method I am working on is this:
Suppose we wanted to compute the gravitational force on the earth from the
known stars and planets. A dauntingly large number of stars must be included
in the calculation, each one contributing a term to the force sum.
In our sum the Andromeda galaxy, which itself consist of billions of stars,
would be included. Since it is so far away it is good enough to treat the
Andromeda galaxy as a single point with a mass equal to the total mass of
the Andromeda galaxy.

More mathematically, since the ratio

                size of box containing Andromeda
      D/r = -------------------------------------
             distance of center of mass from Earth
is so small, we can safely and accurately replace the sum over all stars in
Andromeda with one term at their center of mass.

within the Andromeda galaxy itself, this geometric picture repeats itself:
the stars inside a smaller box can be replaced by their center of mass in
order to compute the gravitational force on a different planet inside the
Andromeda galaxy.

 What I would would like to do is to use Rtree to subdivide space and use
the information on each node. The closer we are to the Andromeda Galaxy,
higher level nodes, with smaller bounding boxes, would be needed.

If the planet is on the same leaf than normal calculation is carried out.

The problem becomes even worse if we want the gravitational force on all
known planets instead of only one.

I hope this clarifies what I need to do. I thought acquiring the members for
each node would be a straightforward matter. But, apparently and
unfortunately , that is not the case...

thank you for the prompt aswer

View this message in context:
Sent from the Boost Geometry mailing list archive at

Geometry list run by mateusz at