Spherical Vector Addition Process

In summary, the conversation discusses the possibility of adding two vectors in spherical coordinates without converting them to Cartesian coordinates first. It is suggested that converting to Cartesian coordinates and back is the easiest method for vector operations. Additionally, the difference between "bound" and "free" vectors is explained, with the example of velocity being a free vector while force is a bound vector. The conversation also touches on adding relative velocities, where the line through which they act does not matter.
  • #1
n00bcake22
21
0
Hello Everyone,

I was just wondering if there was a way to add two vectors that are determined by spherical coordinates (radius, theta, phi). For example, if I have v1 = (5, Pi/4, Pi/2) and v2 = (3, Pi, -Pi/2) is there a way to add these using their respective radii, thetas, and phis or do I HAVE to convert them to Cartesian coordinates first, perform the addition, and then convert back?

Also, if anybody knows, what is the difference between "bound" and "free" vectors? I was looking around and this came up but the description was a bit fuzzy.

Thanks again!
 
Mathematics news on Phys.org
  • #2
hello n00bcake22! :smile:
n00bcake22 said:
I was just wondering if there was a way to add two vectors that are determined by spherical coordinates (radius, theta, phi). For example, if I have v1 = (5, Pi/4, Pi/2) and v2 = (3, Pi, -Pi/2) is there a way to add these using their respective radii, thetas, and phis or do I HAVE to convert them to Cartesian coordinates first, perform the addition, and then convert back?

there's probably a way of doing it once you've found the angle between the two vectors

but converting to Cartesian is the only easy way :smile:
what is the difference between "bound" and "free" vectors?

a free vector is like velocity … you can slide it about anywhere, especially when you want to add it to another vector

a bound vector is like force … you can't slide it about unless you compensate by adding a couple … a bound vector is really a pair, the vector itself and the line of application :wink:
 
  • #3
Thanks tiny-tim!

Converting to Cartesian and back it is! :) Is this the suggested method for all vector operations (dot/cross product, etc.)?

Could you give a slightly more verbose example of “bound” and “free” vectors? I am still a bit confused.

Using your examples, a force A i^ + B j^ + C k^ (bound?) is acting on a point mass M whose velocity is D i^ + E j^ + F k^ (free vector?). I fail to see the difference. :(
 
  • #4
hi n00bcake22! :smile:
n00bcake22 said:
Converting to Cartesian and back it is! :) Is this the suggested method for all vector operations (dot/cross product, etc.)?

yes (except for very simple cases)
Could you give a slightly more verbose example of “bound” and “free” vectors? I am still a bit confused.

Using your examples, a force A i^ + B j^ + C k^ (bound?) is acting on a point mass M whose velocity is D i^ + E j^ + F k^ (free vector?). I fail to see the difference. :(

we don't usually want to add velocities …

although we often add the force on two different points, or the momentum of two different points, we never have any reason to add the velocities of two different points …

but we do add relative velocities, and when we add them the line through which they act doesn't matter, so velocity is a free vector :wink:
 
  • #5


I can confirm that there is indeed a method for adding vectors in spherical coordinates. It is called the Spherical Vector Addition Process (SVAP). This process involves using the law of cosines to calculate the magnitude of the resultant vector and then using the law of sines to calculate the angles of the resultant vector.

In the example given, v1 and v2 can be added using their respective radii, thetas, and phis. This can be done by first finding the components of each vector in Cartesian coordinates, adding them, and then converting the resultant vector back to spherical coordinates.

As for the difference between "bound" and "free" vectors, the term "bound" typically refers to vectors that are confined to a specific coordinate system or frame of reference, while "free" vectors can be moved and applied to different coordinate systems without changing their magnitude or direction. In the context of spherical coordinates, bound vectors are typically used to describe positions or movements within a specific sphere, while free vectors can be used to describe movements or forces outside of that sphere.

I hope this helps clarify the concept for you. Let me know if you have any further questions.
 

1. What is the Spherical Vector Addition Process?

The Spherical Vector Addition Process (SVAP) is a mathematical method used to add multiple vectors in three-dimensional space. It takes into account the direction and magnitude of each vector to determine the resulting vector.

2. How is the SVAP different from traditional vector addition?

The SVAP takes into account the curvature of a spherical surface, while traditional vector addition assumes a flat surface. This makes it more suitable for calculations involving three-dimensional objects or systems.

3. What are the steps involved in the SVAP?

The first step is to convert all vectors into their spherical coordinates. Then, the vectors are added using the law of cosines. The resulting vector is then converted back to Cartesian coordinates for easier interpretation.

4. What are the applications of the SVAP?

The SVAP is commonly used in physics, astronomy, and engineering to accurately calculate the resulting vector of multiple forces or velocities acting on a spherical object. It can also be used in computer graphics to simulate three-dimensional motion.

5. Is the SVAP limited to three-dimensional space?

Yes, the SVAP is specifically designed for use in three-dimensional space. It does not take into account higher dimensions or non-spherical objects.

Similar threads

Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
488
  • Introductory Physics Homework Help
Replies
2
Views
307
Replies
14
Views
1K
Replies
5
Views
4K
  • Calculus and Beyond Homework Help
Replies
7
Views
663
Replies
26
Views
3K
Replies
11
Views
5K
Back
Top