Math/geometry question
Math/geometry question
(OP)
I'm actually ashamed asking you guys for help on this.
I've wasted my entire day trying to solve this and haven't gotten any further yet. I guess I'll have to go back to school and freshen up my math skills.
In attachment you'll find an image of an irregular pentagon with given:
angles: a, b
lengths: H, R, J, C
Requested is the angle marked with a "?" in the drawing.
If I draw it out in NX and define these, I can't move anything else, so I'm pretty sure it "should" be solvable.
I really don't know what I'm missing, any help would be appreciated
I've wasted my entire day trying to solve this and haven't gotten any further yet. I guess I'll have to go back to school and freshen up my math skills.
In attachment you'll find an image of an irregular pentagon with given:
angles: a, b
lengths: H, R, J, C
Requested is the angle marked with a "?" in the drawing.
If I draw it out in NX and define these, I can't move anything else, so I'm pretty sure it "should" be solvable.
I really don't know what I'm missing, any help would be appreciated
NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000





RE: Math/geometry question
The angle that seems 90° (near the "?") is, so you know 3 angles.
NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000
RE: Math/geometry question
RE: Math/geometry question
R=He^ja+Ce^j(a+x-180)+y*e^j(a+x-90)+Je^j(a+x+b-270)
where x is the angle sought and y is the link unknown
From this you write the 2 orthogonal equations in in the variables, x and y.
R=Hcosa+Jcos(a+x-180)+ysin(a+x)+Jcos(a+x+b+90)
0=Hsina+Jsin(a+x-180)-ysin(a+x)+Jsin(a+x+b+90)
The 2 equations should yield the answer, maybe with some sweat.
Good luck!
RE: Math/geometry question
@Zeke, you're also right. It "should" yield the answer, only that's where my problems lie. I can't seem to work it out.
More sweat it is.
Again: All hints are welcome.
NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000
RE: Math/geometry question
Did you try this?
Consider triangle OCD, and polygon OABC.
Forget the trig, try vectors
OC + CD + DO = zero
OA + AB + BC +CD +CO = zero
Too lazy to try myself but looks like it might work.
RE: Math/geometry question
RE: Math/geometry question
so the two unknown angles sum to something known.
then i think you can find the RH lower unknown angle by changing the shape to be a quadrateral, replacing the "dog leg" side with a straight (yes?).
now you've got a quadrateral with three known sides and one known angle.
does that lead anywhere ?
RE: Math/geometry question
Trust me, he already has, and what he has should work (at least I was able to make it work using the same version of NX that Walterke is using). I've attached a file for him to review, but for the rest of you, here's what a sketched solution would look like:
John R. Baker, P.E.
Product 'Evangelist'
Product Engineering Software
Siemens PLM Software Inc.
Industry Sector
Cypress, CA
Siemens PLM:
UG/NX Museum:
To an Engineer, the glass is twice as big as it needs to be.
RE: Math/geometry question
Let
D=C+Jsinb
B=Rsina
A=H-Rcosa
F=sqrt(A^2+B^2)
Then
x=inverse sin(D/F)-inverse tangent(A/B)
Checked against Baker's solution
D=25+160*sin(70)=175.35
B=224*sin(110)=211.75
A=100-224*cos70=23.38
F=213
x=inverse sin(D/F)=55.41 minus inverse tangent(A/B)=6.3
x=55.4-6.3=49.1
compared with Baker's 49.55
RE: Math/geometry question
However, plugging zekeman's equations into excel gives
D 175.3508193
B 210.4911471
A 23.3874879
F 213
asin(D/F) 55.41078923
atan(A/B) 6.340079533
X 49.0707097
----------------------------------------
The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.
RE: Math/geometry question
should be
D=25+160*sin(70)=175.35
B=224*sin(110)=210.49
A=100-224*cos70=23.38
F=211.785
asin(D/F) 55.889
atan(A/B) 6.338
X 49.55
which is now agreed to be the answer
Dgallup,
Thanks for your input
Your F was my old F.
RE: Math/geometry question
I'd like to know how you got to that solution. That's the part I'm having trouble with
Everyone else: thanks for the help!
NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000
RE: Math/geometry question
@Zeke, I'm still curious as to how you did it.
NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000
RE: Math/geometry question
zeke, i'm curious; what's the relationship betwwn the 160 side length, and the angle of the opposite corner ?
similarly the 224 length and the opposite 110deg angle ?
i guessing it's some math relationship for closed polygons (similar to sine rule for triangles) ?
RE: Math/geometry question
"x" appears as the angle of the RH side to the vertical.
RE: Math/geometry question
However, for those interested the trick was to rotate the vector diagram so that the unknown link is horizontal.
Then you write a single equation in x by equating the y ooordinate of the J link end , Jsinb,by adding the y coordinates (from the left of the datum link), thru all the links, C,H and R, fairly straightforward.
RE: Math/geometry question
C 90
H -90-x
R -90-x+180-a=90-a-x
J b-180
so equating the y coordinate of J and setting equal to the sum of the remaining links from the left
-Jsinb=C-Hcosx+Rcos(a+x)
Hcosx-Rcosacosx+Rsinasinx=Jsinb+C
(H-Rcosa)cosx+Rsinasinx=Jsinb+C
Using my notation above
Acosx+Bsinx=D
You can now solve the quadratic equation for sinx,after substituting cosx=sqrt(1-sin^2x) and you are done or
do it using vectors which I find easier to use as follows :
Construct the vector Bi+Aj =F*
where the angle of F* to the x direction is alpha=atan(A/B)
Now rotate F* an angle x and the vector becomes
(Bcosx-Asinx)i +(Acosx+Bsinx)j
Now F maks an angle alha +x to the x axis
And the projection of F* on the y axis is
F*.j=Acosx+Bsinx which is equal to D
From the trigonometry,that angle alpha+x=asin(D/F)
whence x=asinD/F-atan(A/B)
RE: Math/geometry question
RE: Math/geometry question
NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000