Simple coordinate transformation question

Click For Summary
SUMMARY

The discussion centers on the correct formulation of coordinate transformations involving angles and perpendicular distances. Specifically, the equation y' = y cos(φ) is established as correct, while y = y' cos(φ) is deemed incorrect due to the dependency of y' on both y and z. The transformation equations derived include z' = z cos(φ) - y sin(φ) and y' = z sin(φ) + y cos(φ), which accurately represent the relationships between the original and transformed coordinates. The discussion emphasizes the importance of understanding the geometric relationships in the transformation process.

PREREQUISITES
  • Understanding of coordinate systems and transformations
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of geometric principles involving right triangles
  • Basic algebra for solving equations
NEXT STEPS
  • Study coordinate transformation techniques in 2D geometry
  • Learn about the applications of trigonometric identities in transformations
  • Explore the derivation of transformation equations in physics and engineering contexts
  • Investigate the implications of rotation matrices in coordinate transformations
USEFUL FOR

Mathematicians, physicists, engineers, and students studying coordinate transformations and trigonometric applications in geometry.

iScience
Messages
466
Reaction score
5
http://i.imgur.com/MDigPh5.png


if i have my original coordinate (white) and i am transforming this into the red coord. , could someone explain to me why y=y'cos[itex]\phi[/itex] is incorrect and why y'=ycos[itex]\phi[/itex] is correct?
 
Mathematics news on Phys.org
Neither of those is correct unless you are just giving part of the formula- in which case both are correct! y' depends on both y and z and y depends on both y' and z'.

Given any point in the plane, drop perpendiculars from the point to the y and y' axes. The angle at the point is [itex]\phi[/itex], the distance from the point to the foot of the perpendicular to the y' axis is z' and the distance from the point to the foot of the perpendicular to the y-axis is z. Similarly, the distance from the origin to the foot of the perpendicular to the y-axis is y and the distance from the origin to the foot of the perpendicular to the y'- axis is y'.

The distance from the foot of the y-axis to the intersection of that perpendicular is [itex]y tan(\phi)[/itex]. The length of the rest of that perpendicular is the hypotenuse of that right triangle and so is [itex]\frac{z'}{cos(\phi)}[/itex]. Then [itex]z= y tan(\phi)+ \frac{z'}{cos(\phi)}[/itex]. Multiply both sides by [itex]cos(\phi)[/itex] to get [itex]z cos(\phi)= y sin(\phi)+ z'[/itex] so that [itex]z'= z cos(\phi)- y sin(\phi)[/itex]. Similarly, [itex]y'= z sin(\phi)+ y cos(\phi)[/itex].

You can solve those two equations for y and z or simply replace [itex]\phi[/itex] with [itex]-\phi[/itex] (the opposite of "rotating through angle [itex]\phi[/itex]" is "rotating through angle [itex]-\phi[/itex]"). Since [itex]sin(-\phi)= -sin(\phi)[/itex] and [itex]cos(-\phi)= cos(\phi)[/itex] we can just change the sign in front of the sines: [itex]y= -z' sin(\phi)+ y' cos(\phi)[/itex] and [itex]z= z' cos(\phi)+ y' sin(\phi)[/itex].
 
Last edited by a moderator:

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
6K
  • · Replies 7 ·
Replies
7
Views
963