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# Break a vector into component forces2

## Break a vector into component forces

(OP)
I am trying to come up with an Excel solution to breaking up an angular force into its component horizontal and vertical forces. My problem is that the angle of my vector can be from 0 (0 /360 is at the top) to 359 and I want my horizontal and vertical values to be expressed as positive or negative values.
For example:
Force= 35 units at 240 degrees
= in a horizontal direction -17.5 units
= in a vertical direction -30.3 units
or..
-17.5 at 270 degrees and -30.3 at 180 degrees

Can any one help.. it has been way to long since school

### RE: Break a vector into component forces

Ralph
I am a bit confused, do you mean -30.3 at 270 degrees and -17.5 at 180 degrees?
Or are those numbers "guesstimating"?

### RE: Break a vector into component forces

If 0 degrees is vertical and your angles are degrees then the horizontal and vertical components are:

In your example the force is 30 degrees below horizontal, so you have got your components round the wrong way.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

### RE: Break a vector into component forces

(OP)
Thanks AELLC
Those are the values I would expect with a force (vector) of 35 at 240 degrees, Is my math wrong?

### RE: Break a vector into component forces

Your vector is closer to horizontal than vertical; therefore, the horizontal component has to be bigger than the vertical component. Therefore, your x-component = -30.31 and your y-component = -17.50. At 180, your x-component = 0 and your y-component = -35

TTFN
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### RE: Break a vector into component forces

(OP)
Thank you IDS
Perfect for what I wanted to do..

### RE: Break a vector into component forces

darn it.. the last version (r2) is still buggy, returns incorrect for angle = 180 degrees

### RE: Break a vector into component forces

(OP)
Thanks AELLC for trying to help me. Your solution is along the lines I was trying. I have opted to use the solution / formula provided by IDS. It works without a whole lot of if / and / or conditions.
Thanks again

### RE: Break a vector into component forces

Yep I tried IDS's method - the sign is automatic. Much cleaner.

### RE: Break a vector into component forces

A couple of other points:

If you need to go the other way (find the resultant angle from the vertical and horizontal components) use the ATAN2() function, as this will give a result over the complete circle, rather than just 0 to pi radians.

To convert degrees to radians use the Radians() function, or multiply by PI()/180, rather than typing in an approximation to pi.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

### RE: Break a vector into component forces

(OP)
Thanks IDS for your help, very much appreciated.
Another scenario.. To calculate the tension in a sling used at an angle
Vertical known
Angle of hypotenuse (will always be between 0 horizontal and 90 vertical) known

Want to know value of horizontal and hypotenuse.

### RE: Break a vector into component forces

(OP)
I need help again.
I am not understanding how to use IRstuff latest formula Fv/atan(theta) to solve my problem.
A1=Vertical force
A2=Angle of force (hypotenuse of right angle triangle)
With A1 and A2 known is there enough information to determine A3 and A4
A3=______________ horizontal force
A4=______________ amplitude (length) of hypotenuse)

This to calculate the tension on a sling that is used in a none vertical position.

Thank you again for all the previous help

### RE: Break a vector into component forces

(OP)
Thanks again to all
Resolved my own issue after much frustration. The final solution to solve for the long side of a triangle when the angle and one side is known is =A1/ SIN(A2*(3.1416 / 180)). My problem was that the numbers were not coming out right.. It seems that Excel uses radians and not degrees .. thus the continuing errors until I added the (3.1416 / 180) to convert my degrees to radians.

### RE: Break a vector into component forces

Doug pointed that out in his first posting here, since he used COS(RADIANS(Angle)) and further mentioned that the RADIAN() function avoids the need to use or calculate pi and dividing by 180 degrees. Note, also, that Excel has a PI() function, so there is no need to explicitly calculate pi.

TTFN
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