Quote Originally Posted by aratti View Post
Inverse sin is not possible with PBP but inverse tan yes (I am sure to have seen an example (ATN2) given by Darell sometime ago).

In this case you can just manipulate a bit your equation to obtain the tangent, since:

SIN = PERP./HYP. (where: (v-v0) = PERP. and (v1-v0) = HYP.)

TAN= PERP./BASE (where: (v-v0) = PERP. and SQR((v1-v0)^2 - (v-v0) ^2) = BASE)

so:

ATN2 ((v-v0)/SQR((v1-v0)^2 - (v-v0) ^2))

should give you your angle (I don't remember if in degrees or in radiants)

Edited: See also this link:http://www.picbasic.co.uk/forum/show...ighlight=angle

Al.
So over a year and a half has passed and suddenly i am back in the same position i was in before. Due to personal reasons i had to drop out of the college course i was doing and now i am back on it again... so ive picked up the demo of the new PBP and am finally about to try this out!

The first thing i notice is an issue with the equation "ATN2 ((v-v0)/SQR((v1-v0)^2 - (v-v0) ^2))"
As far as i can find ATN2 does not exist in the manual and throws back errors in the compiler! After looking into this I believe you meant to use just ATN? the second question i would have if this is correct would be the positioning of the ATN command. As i cannot get to my practical equipment until later this week would you be able to read this and see if i am on the correct road please:

Perp = v - v0
Base = SQR((v1-v0)^2 - (v-v0) ^2) = BASE)
Rad = Perp/Base

Any guidance would be greatly appreciated! It can be hard to get the brain to start going over all these things again!