Interesting. It looks like the radius computed from the polygon perimeter is the Authalic Radius, which is counting the length of the sides of the pixels that form the boundary.
In[89]:= (1 /. ComponentMeasurements[img, "AuthalicRadius"])*2
Out[89]= 363.41
If you compute the diameter instead based on the ConvexPerimeterLength, it gets a bit closer to the measurements given by the Length and Width properties that describe the best-fit ellipse. The points along the convex hull don't hug the pixel elements closely and look a bit closer to what you'd expect of a circle-like polygon around the component.
In[88]:= (1 /. ComponentMeasurements[img, "ConvexPerimeterLength"])/Pi
Out[88]= 346.648
You can see the difference between the polygon perimeter and the convex perimeter if you use a smaller circle and plot the points along the boundary. In the example here, the blue points are the perimeter positions of the boundary pixels, while the orange points are the vertices of the convex hull.
img = ColorNegate@ColorConvert[Image@Graphics[Disk[], ImageSize -> {50, 50}], "Grayscale"];
ListLinePlot[{(1 /. ComponentMeasurements[img, "PerimeterPositions"])[[1]],
(1 /. ComponentMeasurements[img, "ConvexVertices"])},
PlotRange -> All, Joined -> False]
Interestingly, computing the area and deriving the diameter from that gets closest to the ellipse parameters computed by Length and Width.
In[15]:= Sqrt[(1 /. ComponentMeasurements[img, "Area"])/Pi]*2
Out[15]= 345.658
In[13]:= ComponentMeasurements[img, #] & /@ {"Length", "Width"}
Out[13]= {{1 -> 345.63}, {1 -> 345.502}}
Unfortunately, the convex perimeter isn't going to help you if you are looking to calculate the perimeter of a non-convex object in the image. You might be stuck with the PerimeterLength measurement for those kinds of shapes.
