Knead
February 12, 2017, 7:10pm
1

https://docs.mcneel.com/rhino/6/help/en-us/commands/matchsrf.htm

Match Surface Options

Curvature __ percent
Curvature matching, in percent of the radius of curvature.

I’m trying to understand what the percent option means.

For example: curvature is default 0.05 percent

(A) Radius = 20 mm : The CURVATURE VALUE ( 1/Radius) would be 0.05
(B) Radius = 50 mm : The CURVATURE VALUE ( 1/Radius) would be 0.02

0.05 - 0.02 = 0.03 < default 0.05 percent
Thus, The two surfaces would be 100% curvature continuous?.. No?

pascal
(Pascal Golay)
February 12, 2017, 11:54pm
2
Hi Knead - the % curvature refers, admittedly ambiguously at best here, to radius if curvature.

-Pascal

jim
February 13, 2017, 12:09am
3
Pascal,
I’m I correct in saying that all these settings are meaningless unless you have the “refine” option set.
In other words, if you don;t have “refine” set you are going to get the same result no matter what numbers you fill in the boxes.

Knead
February 13, 2017, 12:31am
4
It is very difficult to understand what it means.

pascal
(Pascal Golay)
February 13, 2017, 12:38am
5
Hi jim - I believe that is correct, yes. I have not actually tested though, in all these years of using it… I’ll see if I can come up with a meaningful test case.

-Pascal

pascal
(Pascal Golay)
February 13, 2017, 12:43am
6
Hi Knead - The radius of curvature of the surfaces should be the same to within that % at points along the matching edges.

-Pascal

Knead
February 13, 2017, 1:48am
7
for example:
(A) Radius = 20 mm
(B) Radius = 50 mm
20 mm VS 50 mm

Thus, 40% = Maximum difference in curvature between surfaces.

Does the percent mean to compare the physical size of the radius (20mm and 50mm), not the values of the curvature (1/radius = 0.05 and 0.02)? right?

chuck
(Chuck Welsh)
February 13, 2017, 3:59am
8
The percent difference between 0.02 and 0.05 would be
100*(0.05 - 0.02)/(some measure of the general size of 0.02 and 0.05), for example
100*(0.05-0.02)/0.05 = 60%. Whether you use, the radius or the curvature the answer is roughly the same.

Knead
February 13, 2017, 6:39am
9

MatchSrf_Tolerance_percent_Knead_1st.3dm (354.2 KB)

in attached file:
(Blue) radius = 81.5069601 mm
(Red) radius = 81.4166767 mm

100 * (81.5069601 - 81.4166767) / 81.5069601 = 0.1107677183509633 %
When comparing Radius size in%, Matching tolerance exceeds 0.05%.

or

1 / 81.5069601 = 0.012268890887025
1 / 81.4166767 = 0.0122824959275205
100 * (0.0122824959275205 - 0.012268890887025) / 0.0122824959275205 = 0.1107677183512528 %
When comparing curvature in%, Matching tolerance exceeds 0.05%.

Thus, the radius or curvature of the matching surfaces have not the same within tolerance 0.05%.
What’s wrong with this?

Knead
February 13, 2017, 12:10pm
10
The Curvature Difference Method.

Curvature Deviation = Curvature (Blue) - Curvature (Red).
Curvature is 1 / Radius.
Curvature Deviation = 1 / Radius (Blue) - 1 / Radius (Red)

In the attached file,
Curvature Deviation = 1 / 81.5069601 (Blue) - 1 / 81.4166767 (Red)
= 0.012268890887025 (Blue) - 0.0122824959275205 (Red)
= 0.00001360504049546093
= 0.00001360504049546093 * 100
= 0.0013605040495461 percent

Thus, 0.0013605040495461% is the same within tolerance 0.05%
Right? !

chuck
(Chuck Welsh)
February 13, 2017, 4:48pm
11
Sorry, I was just pointing out that in your first example, with 20 and 50, your calculation was incorrect.

Knead
February 14, 2017, 2:54am
12

chuck:

Sorry, I was just pointing out that in your first example, with 20 and 50, your calculation was incorrect.

in attached file:
(Blue) radius = 81.5069601 mm
(Red) radius = 81.4166767 mm

100 * (81.5069601 - 81.4166767) / 81.5069601 = 0.1107677183509633%

I thought the result of the matching surface should be within tolerance.

0.1107677183509633% > 0.05%

But, When comparing Radius size in%, exceeds 0.05%.
What’s wrong with this?

chuck
(Chuck Welsh)
February 14, 2017, 7:30am
13
Is your problem in the numbers, or is there something wrong with your model?

Knead
February 14, 2017, 4:51pm
14
Command: _MatchSrf
and
Check Refine match option.
Input Curvature 0.05 % percent

But, The radius of curvature of the surfaces was not the same to within that 0.05% at points along the matching edges.

chuck
(Chuck Welsh)
February 14, 2017, 5:07pm
15
Yeah, something isn’t right. We’ll look into it. I’m not sure this is a very good measure of accuracy.