Plot3D
, because it samples only in the domain-plane (the xz
-plane in the OP), has trouble resolving surfaces with tangents perpendicular to the domain-plane. For functions whose gradients are not very large, Plot3D
is usually superior to ContourPlot3D
, but in this case, I think ContourPlot3D
works better.
Even with the suggestions of PlotPoints -> 100
and MaxRecursion -> 5
, there are still some tiny but visible glitches in the surface mesh:
Plot3D[(((-0.0023 x^2 + 0.525 x + 1)^2) *
((1 - (((z)^2)/((12 - ((0.0006 (180 - x)^2 - 0.1118*(180 - x) + 1)))^2)))^(1/2))),
{x, 0, 180}, {z, 0, 16},
PlotPoints -> 100, MaxRecursion -> 5, AspectRatio -> Automatic,
AxesLabel -> Automatic, ImageSize -> Medium,
ViewPoint -> {3.215347035920252`, 0.7760554973318896`, 0.7136394773692102`}
] // AbsoluteTiming

It is also necessary to square both sides to get rid of the square root. (Sampling outside its domain and possibly rounding errors result in negative numbers under the radical, which causes ContourPlot3D
trouble in resolving the surface.)
ContourPlot3D[y^2 == (((-0.0023 x^2 + 0.525 x + 1)^2)^2 *
((1 - (((z)^2)/((12 - ((0.0006 (180 - x)^2 - 0.1118*(180 - x) + 1)))^2))))),
{x, 0, 180}, {z, 0, 16}, {y, 0, 958.5},
MeshFunctions -> {#1 &, #2 &},
MaxRecursion -> 2, BoxRatios -> {1., 1., 0.4}, ImageSize -> Medium,
ViewPoint -> {3.215347035920252`, 0.7760554973318896`, 0.7136394773692102`}
] // AbsoluteTiming

Plot3D[(1 + 0.525 x - 0.0023 x^2)^2 (1 - z^2/(11 + 0.1118 (180 - x) - 0.0006 (180 - x)^2)^2)^0.5, {x, 0, 180}, {z, 0, 17}, AspectRatio -> Automatic, PlotPoints -> 100]
$\endgroup$PlotPoints
recommend that you useMaxRecursion->5
$\endgroup$