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petercorke / robotics-toolbox-matlab

Robotics Toolbox for MATLAB
GNU Lesser General Public License v2.1
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lspb (Linear Segment with Parabolic Blend) problem #97

Closed A6GH closed 3 years ago

A6GH commented 3 years ago

s = lspb(s0, sf, m) function call for bounds s0 = 0, sf = 1 and m=36 steps returns the last element of s to be greater than 1. Matlab version R2020a.

petercorke commented 3 years ago

Odd, it's fine for me, using 20b I get:

>> tg = lspb(0, 1, 36)

tg =

         0
    0.0018
    0.0073
    0.0165
    0.0294
    0.0459
    0.0661
    0.0900
    0.1176
    0.1488
    0.1837
    0.2222
    0.2643
    0.3071
    0.3500
    0.3929
    0.4357
    0.4786
    0.5214
    0.5643
    0.6071
    0.6500
    0.6929
    0.7357
    0.7778
    0.8163
    0.8512
    0.8824
    0.9100
    0.9339
    0.9541
    0.9706
    0.9835
    0.9927
    0.9982
    1.0000
A6GH commented 3 years ago

Odd, it's fine for me, using 20b I get:

>> tg = lspb(0, 1, 36)

tg =

         0
    0.0018
    0.0073
    0.0165
    0.0294
    0.0459
    0.0661
    0.0900
    0.1176
    0.1488
    0.1837
    0.2222
    0.2643
    0.3071
    0.3500
    0.3929
    0.4357
    0.4786
    0.5214
    0.5643
    0.6071
    0.6500
    0.6929
    0.7357
    0.7778
    0.8163
    0.8512
    0.8824
    0.9100
    0.9339
    0.9541
    0.9706
    0.9835
    0.9927
    0.9982
    1.0000

It probably has something to do with floating point precision. The last number generated is not exactly 1. Try tg(36) == 1. lspb works fine by itself, but if you call ctraj (which uses lspb) on 36 points, it will report an error when calling trinterp.

petercorke commented 3 years ago

Definitely a FP issue, it's over by 4 eps which is small in the scheme of things. It's evaluating a quadratic. Quick and dirty fix would be to clip the value. I'll add this to the be fixed list, but it won't happen any time soon. Thanks.