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3D Vector 2

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sbozy25

Mechanical
Joined
Jun 23, 2005
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395
Location
US
Ok... I feel so incredibly stupid for asking this, but I as well as my electrical engineer are stumped.

I am working on an issue regarding side loading in compression springs, and I am analyzing the data our instron tester is giving us. From this data I will be able to tell at what position in 3D space the test force is occuring and translate it back to a side load and moment. However, I for some reason am getting answers that do not make sense.

Ok... Here is the problem...

I have three positions of the spring, top side P1, bottom side P2, and test height P3. I have the (x,y,z) coordinates for P1 and P2 but only the z for P3. P3 is on the same line as P1 and P2....

so correct me if I am wrong in this assumption....

Vector A: (P1x-P2x)i + (P1y-P2y)j + (P1z-P2z)k
Vector B: (P3x-P2x)i + (P3y-P2y)j + (P3z-P2z)k
Vector C: (P1x-P3x)i + (P1y-P3y)j + (P1z-P3z)k

Vector A = Vector B + Vector C Is this not correct?

The problem is P3x and P3y are un known and I need to solve for them, but in the above equations they end up cancelling out as thogh they are zero... I don't know, perhaps I am over analyzing this... can anyone provide insight? I can give actual data if needed.... Thanks
 
You have more unknowns then independent equations.

-Reidh
 
I think your equations are completely wrong.

For instance, if P3z = P2z then it should say that P3x=P2x and P3y=P2y

P3x=(p3z-p2z)*(p1x-p2x)/(p1z-p2z)+p2x

etc

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
 
The way you do this is, since p3 is inline,and A=P2-P1 then the vector C=P3-P1 is equal to kA where k is a constant and equal to the ratio of z3-z1 to z2-z1, so
k=(z3-z1)/(z2-z1).
To facilitate matters you should change your coordinate system so that P1 is at 0,0,0. Then all the vector components will also be the x-y-z coordinates.
 
Hi,
I'm with Reidh. The Chasles theorem A=B+C is not enough, you need another relation between coordinates if you want to solve for TWO unknowns deterministically (two unknowns call two equations). If you don't have this relation, then there are infinite pairs (P3x;P3y) which can satisfy the only equation you have, so the problem is indeterminated by 1-infinity.
For example, Zekeman's additional relation can be a solution, but it's valid under the hypothesis of linearity (is this hypothesis realistic in your case?). In fact, if P3 is in line with P1 and P2 initially (zero-deflection), it may not be the case under deflection (the axis line of the spring remains linear during the deflection? I don't think so, I'd rather think it's parabolic from P2 to P3 and linear from P3 to P1). In fact, if I'm not totally drunk, under Zekeman's hypothesis y=mz and the line of vector P1-P3 should pass through P2, which is very hardly the case.

Regards
 
Thanks everyone for the help. Unfortunately, zekemen I can not change my coordinates for a (0,0,0) situation, because that would defeat my entire purpose.

I have the following points...

P1: (-.076, .135, 0)
P2: (X2, Y2, 1.529)
P3: (.188, -.045, 7.917)

It is X2 and Y2 that I need to find, because from there I will be able to find the perpindicular force vector that creates my loading issue...
 
Yes you have, I was just putting that data up so zekemen would see that I couldn't change my coords.
 
Quote"Thanks everyone for the help. Unfortunately, zekemen I can not change my coordinates for a (0,0,0) situation, because that would defeat my entire purpose.

I have the following points...

P1: (-.076, .135, 0)
P2: (X2, Y2, 1.529)
P3: (.188, -.045, 7.917)

It is X2 and Y2 that I need to find, because from there I will be able to find the perpindicular force vector that creates my loading issue..."

Then to get P2 you simply add -.076,.135,0 to kA
k=1.529/7.917=0.193
and therefore
P3-P1=.188+.076,-.045-.135,7.917=.264,-.180,7.917
and P2=k(P3-P1)+P1=0.193*(.264,-.180,7.91)+-.076,.135,0

P2=-.02505,.100,1.529
 
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