# Evaluate 3D Shape Descriptor

I'm trying to create my own 3d shape descriptor, the idea is that how I may evaluate how much my descriptor is well and good?

What I checked is that they evaluate descriptors through shape matching, shape retrieval, etc..

But my case now is before applying my descriptor to an application, though, Descriptor could be well if it satisfies properties like being invariant to transformations (scale, rotation, translation).

How much this is accurate, and if we suppose this, that I already trying to implement, my case is that I have feature vector for descriptor a for main model and same descriptor for the scaled object for example, How should I verify that it's invariant to scale or not? what kind of equation applied for example!

• Can't you compute the descriptor for the object, scale and compute again? – nicoguaro Jan 29 '20 at 13:05
• InterestPoints(M) = [866 0 2 135 137 138 141 142 162] and InterestPoints(S) = [866 0 1 2 4 5 6 7 8 4 175 176] the results are abbreviated,the for each vector of IP I get the descriptors, let's assume it the mean curvature descriptor of these data like what follows: Mean_Curvature_Main = [0.75, 0.0, 0.0, 0.263, 0.933, 0.994, 0.994,0.217...] Mean_Curvature_Scale = [0.736, 0.0, 0.241, 0.24 ,0.24, 0.018, 0.302....] What I the proper way to compare the results in order to say whether this descriptor is invariant under scaling or not!! @nicoguaro this what you mean? – R.K Jan 29 '20 at 13:22
• I don't think so, curvature is not dimensionless. So, I doubt it is scale invariant. – nicoguaro Jan 29 '20 at 13:28
• it's a single case, I'm trying different descriptors like, mean curvature, Shape Index, Curvature Index.. and then make a formula to obtain a new descriptor, so what I'm trying to achieve is to show how each descriptor is valuable, in order to figure out whether mine is better or not! I'm trying , scaling, translation, rotation, etc.. Was I be able to clarify what I'm trying to do? :) – R.K Jan 29 '20 at 13:35
• @nicoguaro I think OP needs to make the geometrical descriptors dimensionless in order to show they are invariant under linear and isotropic transformations. Yes, Gaussian curvature unit is $\mathrm{m}^{-2}$, but OP could for example normalize over the surface area of the 3D shape and that number should be invariant under linear and isotropic transformations. – Alone Programmer Jan 29 '20 at 13:53

• The idea Sir, is that I'm working on feature vectors not histogram-ing, weird, but till now I wasn't able to get how complicated descriptors work, so I tend for simple ones. So, assuming I have feature descriptor a having at each point a value, and same descriptor, but for another model(scaled, translated, transformed...) b my question how I may compare(match) these models depending on their descriptors, what should be the metric to say that this descriptor is invariant under such transformation? – R.K Feb 5 '20 at 10:46