# Tag Info

0

"...re-adding the horizontal constraint" "My constraints and components are exactly the same." The statement above is not true - in one case you the dimension is measuring a distance to the right of the origin, and in the other, the distance to the left. When entering a dimension in SolidWorks, there is the option to flip the direction ...

1

One way would be to give the angle with the horizontal, instead of providing the 5mm length. The problem in this case is that the angle in most cases will not be a round number. However, you might be able to use a function in the dimension (e.g. atan(3/5)).

0

hmm. here's a shot in the dark. might require relatively simple surface. this is crude, the are much more sophisticated ways. research metric or distance for 3D shapes. (1) In a 3D modeling tool, fit a simplified (curvature never too small) convex dummy surface to a typical data surface. Shrink it maybe 50% in a 3D tool, so it is always "inside" ...

0

You can superpose the black and white pixelated photos of your shapes assigning a grayscale value of $1/n * (1/255) \ ,$ per shape. you will get an image with a black core and shading to gray and feathering to white. Many photo editing softwares allow you to define a border based on the gray intensity. In your case, it is 127-128. If some parts of your ...

0

$L$ = center line distance of the pin-hole assembly, = 235 $D_H$ = diameter of the hole, = 46 $d_H$ = tolerance of the hole, = +0.004, -0.000 $D_P$ = diameter of the pin, = 46 $d_P$ = tolerance of the pin, = +0.000, -0.004 $A = L + 2(D_H/2 + d_H/2) = L + D_H + d_H = 235 + 46 + 0.004$ $B = L - 2(D_H/2 + d_H/2) = L - D_H - d_H = 235 - 46 - 0.000$ $C = L = 235$...

Top 50 recent answers are included