explanation of hit/vector under linearopen scan

[toffield]

Using 3.5mr2.

I created a linearopen scan. I wanted to find the min of the scanned points I took. When it reported the min distance out I couldn't find it under the hit/vector

Could someone explain the following.

hit/vector,-.5132,-0.0939,0.017,0,-0.0132289,-0.9999125,-.5132,-.0939,0.012,t=.0005

I know the first 6 location are xyzijk, the next 3 seem to be xyz and then the deviation.


I do not have a model. Why when I report the hit value in the z it reports .012, not .017.

Any help is greatly appreciated.

Teresa

[Duckhunter]

explaination of scan point

hit/vector,-.5132(nom x),-0.0939(nom y),-,0.017(nom z),0(nom i),-0.0132289(nom j),-0.9999125(nom k),-.5132(meas x),-.0939(meas y),0.012(meas z), t=.0005(total deviation(t)

[toffield]

Tim,

Thanks for the explanation. That's what I thought it was, now I have another question. Where does it get the nominal data from? I do not have a model.

Again, thanks for your help.

Teresa

[bw_bob]

Tim,

Thanks for the explanation. That's what I thought it was, now I have another question. Where does it get the nominal data from? I do not have a model.

Again, thanks for your help.

Teresa

When you set up your alignment it gets it from that.

[toffield]

bw bob,

If I enter the vectors I need for the surface under "INITVEC" shouldn't those be the vectors I see with the nominal hit vectors? It's a 17 degree angled surface.

The vectors I see are 0, -.0132289, -0.9999125 and it should be .2923075, 0, -.9563244.

This is scanning a datum surface that is .999 long and scanning along the Yaxis, in the -Zaxis. This is the first run through. When it is completely set up within .0005" would it then have the correct vectors?

Thanks for your help!

Teresa