[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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