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如何使用 pcl 计算自旋图像描述符

[英]How to compute spin image descriptors with pcl

I have a point cloud using the c++ pcl library and I would like to compute the spin image descriptor for each point.我有一个使用 c++ pcl库的点云,我想为每个点计算自旋图像描述符。 I have tried with the implemanted SpinImageEstimation class in pcl but I get a histogram way larger that it should be.我已经尝试在pcl中使用已实现的SpinImageEstimation class 但我得到的直方图比它应该的要大得多。 For example with a spin image width of 8 , I get a histogram of size 153 .例如,旋转图像宽度为8时,我得到大小为153的直方图。 But I would like to have the 8x8 spin image.但我想要8x8旋转图像。

Do you know how I should proceed for this?你知道我应该怎么做吗?

ps: for reference on the spin image descriptor, https://www.sciencedirect.com/science/article/pii/S0262885698000742 ps:自旋图像描述符参考, https://www.sciencedirect.com/science/article/pii/S0262885698000742

I am not sure why you think the spin image is 8x8.我不确定您为什么认为自旋图像是 8x8。 See equation (2) in the paper: the spin image is rectangular because the points can lie on the positive or negative side of the normal.见论文中的等式(2) :自旋图像是矩形的,因为点可以位于法线的正侧或负侧。 If you set the spin image width/bin count to 8, the image will be (2*8+1)x(8+1), which equals 153, the histogram size.如果将旋转图像宽度/bin 计数设置为 8,则图像将为 (2*8+1)x(8+1),等于 153,即直方图大小。

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