Trigonometry Formulas

[Pisces]

Originally Posted by JMV

Sorry Pisces,
I must be a very dumm one, I don't get a damn thing on your before last thread...
Thanks for trying however. JMV

That's too bad. Wasn't sure how much background info/ formula derivation you were after. No, no, I don't want to hear that!. Dummies go sit in a car that is about to get crashed on purpous. I may not be the best in explaining complicated stuff. (besides, what's a math discussion worth thesedays without a drawing to make it clear) Don't give up yet. ... But why don't you come up to the chalkboard, .... here's a piece of chalk, ... and show where you get lost.

You wanted formulas for use with a scientific calculator. I supposed you knew how to translate those formulas into keypresses for your own specific brand of calculator. As each brand has it's own way of organising math functions with certain buttons, and how to acces them. I could help with a Casio probably (I have a "fx-82SX Fraction", pretty ancient by now) but some calculators are different in their use. Is that the problem?


p.s. my thread? You must mean my 'post' or my message. You are the one that started this 'thread' with your original message. Sorry, I'm a semantic nitpicker aswel sometimes. And wanted to make sure you didn't mean someplace else.

[JMV]

Hello Pisces,

Thank you for helping. There is an old saying : "when the student is ready, the master appears " in french, if you can understand, " Quand l'eleve est pret, le Maitre apparait " Ok, obviously the time is not there yet for me.... You showed up but it doesn' seem I'm ready...But let's leave it there with phylosophy and mysticism.
I have a Casio fx-83ES. The problem is not with using the Calculator, and after all I've got the manuals, so this is another topic ( How to use your calc machine ) this is not the real issue.
The real issue for me is to get all "navigation" ( only ) trig formulas applied for Sh3. With down to earth concrete examples.
What I probably miss, if I have a formula under my eyes, is a down to earth example for such things as SPEED, DISTANCE, AOB calculation. Basically I can't choose what formula applies to what situation.
For example you write :
side A/sin(angle opposite side A)=side B/sin(angle opposite side B)=side C/sin(angle opposite side C). I mean, what gives ? To what particular situation does it apply?
And what does it apply to ? Are we searching for an angle or for the length of one of the sides ??
So from what I understand :
If you look for the length of the adjacent side of a triangle, knowing the value of the angle which has this side as adjacent, and knowing the length af the opposite side of that angle, you can know the length of the other side, with :
Tan (A) = Length of opp. side
--------------------
Length of adj. side
If you look for the value of an angle, having the length of the opposite side of that angle, and the adjacent angle,
Sin (A) = Length of opp. side applies
-------------------
Length of hypotenuse
Now, Cos (A) = Length of adj. side What do you get ? the length of one of
------------------- the other sides or the value of an
Length of hypotenuse angle, and if yes, which one ?

You can see now that I'm very thick on those trig things, but I suppose, from what I've seen, that I'm not the only one... And I'm not the only one that would like to get very basic applied trig solutions for SAH3, whithout daring ask or look too stupid. So I making the sacrifice.Or may be I sould take some private trig 50 Euro/hour courses...

Thanks for any help, and to you Pisces, please be patient and methodical to me
poor numb on that subject. Don't ever surrender trying. Thanks to you all .
Thanks for your help, if you have the patience.
- JMV

[Pisces]

@JMV

Sorry for not responding right away. Some things got in the way.

Well, that formula:

side A/sin(angle opposite of side A) =side B/sin(angle opposite of side B) =side C/sin(angle opposite of side C)

states the relationship between sides and angles in triangles. It is something that should have been taught right after you learned about what sin, cos and tan means. The above is known as the "law of sines" and applies to ALL (flat plane) triangles in ANY shape or size. (I'm never sure if it's called "rule" or "law"; anyway there is also a "law of cosines" and a "law of tangents") ... Draw 3 random dots on a piece of paper, draw lines between them, measure the sides and angles accurately, and you'll see those 3 divisions are all equal. ... But you are right, if you don't know how to apply it to real situations, it's useless. But we'll fix that soon.

Let's start now. I suggested the intercepting problem. Have a look at the following drawing. (very dirty, sorry)


In it you see a moving target, and your (yo)U-boot intending to meet at some place. The target doesn't know ofcourse, but he'll play along for this explanation. The target is moving along some course with a certain speed indicated by his arrow. Far away from him, you received his location,course and speed by radio and have drawn it on the map. You see that the target is at the end of certain bearing-line (thick black line). Simply a line 'as the crow flies'. You are trying to work out which course and speed to set to meet him at the quickest possible conveniance.

I hope you realise that setting course directly to his current position (along the drawn bearing-line) is not the way to go. By the time you have reached his position, he will be further along his course, and likely far outside of your sensor range (hydrophone). This is because of the sideways component of his speed (dark red thin arrow perpendicular to bearing line). He is going to close-in on you a bit because there is some speed component inwards, but mostly sideways. In other situations the target may be directed away or go to the right depending on his course and position. You need to match this 'sideways' speed before you can invest the rest of your speed in closing the distance (along that bearing-line). That is why I have drawn (thin dark blue) arrow components next to your intended speed and course arrow. The target's red, and your blue, thin arrows (perpendicular to the bearing line) are of equal size. This keeps the target at a constant direction relative to you. He won't appear to move sideways if could magically see him with your periscope. The bearing line to him wil stay in the same direction as you both move. But now only the target is the one that is doing al the work to get closer. That's going to take a long time, and you don't want to risk he's making a course change before you get him. Since you will have excess speed compared to that sideways speed (your thick black arrow is longer) you can use it in the direction directly towards the target(and that doesn't change along the road). The other components (along the bearing line) make you close-in to one another. Unfortunately their lenghts aren't as simple as subtracting 'perpendicular' from 'full speed' because the angles 'skew' the arrow lengths. (Blame Pythagoras for that)

So, there's your problem: You have 2 speeds (you and the target) that must 'butt heads'. You know one (the target) is aimed at some angle (AOB) to the bearingline. You know his speed and know (or choose) your speed. How do you aim your speed against the bearing line? This problem can be represented with an irregularly shaped triangle where the sides are speeds!

You said you (only) know that sin=opposite side/hypotenusa; cos=adjacent side/hypotenusa and tan=opposite side/adjacent side. That's right ... BUT those only apply to triangles that have a 90 degree angle in them (right-angled triangle). Those cannot be used directly for irregularly shaped triangles with 3 'any number of degrees'-angles (like a 3-62-115 degree triangle, or even 60-60-60). Do you get that? This is important because nature almost never gives us problems to solve that have nice right-angled triangles. Most of the time you get irregular stuff. Also in the navigating business (as we can see with the courses in the picture above).

But that's no problem. Because we can introduce them ourselves. We simply divide the irregular triangle in two 'right-angled' triangles, by placing a perpendicular line underneath one of the corners, and use it as an 'internal side' for intermediate calculations. (did a lightbulb flash on yet?)

Using only that "sin=opposite/hypotenusa", we first look at the (dark red) right-angled triangle of the target:

Hypotenusa= target_speed
Opposite side= unknown perpendicular_component
Angle= AOB

So: perpendicular_component= target_speed * sin(AOB)

(Note: this is our minumum speed! If that is too much for your uboot, forget about the target!)

Now we are going to match that perpendicular component with your (dark blue) speed triangle:

Again, using only that "sin=opposite/hypotenusa":

Hypotenusa = your_speed;
Opposite side= now known 'perpendicular_component'
Angle= unkown 'intercept_angle'

So: sin(intercept_angle) = perpendicular_component / your_speed

but in a more usefull way:

intercept_angle= arcsin(perpendicular_component / your_speed)

Great, problem solved, you now know where to turn to: namely, the intercept_angle to the left of the bearingline. If the target is moving away though, your 'along'-speedcomponent may not be enough to overtake his 'along'-speedcomponent. So you need to check that too.

Now let's put these two sine-equations into one single equation and see what we get (for curiousity sake):

for the target: perpendicular_component= target_speed * sin (AOB)

modify your's first; it is the same as: perpendicular_component= your_speed * sin (intercept_angle)

Both perpendicular_components are equal so:

target_speed * sin(AOB) = your_speed * sin(intercept_angle)

Does that ring a bell? No? Maybe the following will, it's the same as above:

target_speed / sin(intercept_angle) = your_speed / sin (AOB)

We just proved the 'Law of Sines' !!!!

Then you might ask, where 's the 3rd part? Well you only need 2 sides for an equation symbol (=) to work. We haven't had a need for the 3rd sofar. But if you look at the drawing again, you see those components along the bearing line. That's the 3rd side. Added up together they show how fast you close to one another. If you know the 3rd angle (and you do!, because there are only 180 degrees in any triangle, and 2 of the 3 angles are now known, so 3rd angle= 180- AOB - intercept_angle) you can use either one of the first 2 'Law of Sines'-parts to calculate the closing speed, as:

target_speed/ sin(intercept_angle) = closing_speed/ sin(180 - AOB - intercept_angle)

or

your_speed / sin (AOB) = closing_speed/ sin(180 - AOB -intercept_angle)

But: sin(A)=sin(180 - A) because of symmetry reasons, so you can just use sin(AOB+Intercept_angle) instead in those formula's.

I'll straigthen those two out so you can see how to calculate closing_speed:

closing_speed = target_speed * sin(AOB + intercept_angle) / sin(intercept_angle)

and

closing_speed = your_speed * sin( AOB + intercept_angle) / sin (AOB)

Now the time_to_meetingpoint in hours is simply the length of the bearing line (initial distance converted to nautical miles) divided by this speed. (in knots naturally)

That's the end of math class for now. For next weeks assignment I want you to solve 'Fermat's last theorem'.

No, kidding asside, I tried to make every step in the thought process as basic as possible. But if there is still something not clear enough let me know.

[JMV]

Dear Pisces,

Thank you so much for answering me in such an extensive and detailed manner.
As you can understand, I've got to get in the deep of your point, and I can't answer right now. It might take a bit of time before I answer back , but after all we're not catching a train here ( only a sub... ) :p . So, take your time too.
In the meantime, I stumbled on this recently :
http://www.hnsa.org/doc/tdc/index.htm ( for our matter see Section 1 )
The printing of the formulas is pretty bad though !
http://www.mathdaily.com/lessons/Trigonometric_function
http://www.math.com/homeworkhelp/Trigonometry.html
http://www.angelfire.com/nt/navtrig/index.html
http://home.alltel.net/okrebs/index9.html ( You probably know all that already... I have a few more sites if you need)


After having gathered 2.39 Gb ( in 2172 files and 391 folders ) of internet sites, pdf's, and other docs, I'm feeling for toughening my realism one or two notches up... All my docs and sites are availiable to you, if you're looking for anything, buzz me. I'm not saying I have every docs and net sites on record, but I'm skimming the forum methodically...
Thanks for all again.
JMV

[JMV]

Pisces,

Just saw a few videos on Fermat.... I have enough with what I have there for a while, "Heaven can wait" as they say.
Are you a mathematician yourself ? You seem incredibly fluent with all that. JMV

[Pisces]

Originally Posted by JMV

Pisces,

....
Are you a mathematician yourself ? You seem incredibly fluent with all that.

Nah, just payed more than good attention in math- and physics-class. Classical-nerd-syndrome. And that was just medium level education. Slowly worked myself upto aeronautical engineering college level. But I wouldn't stand a chance at the university level. Too many goofups and rash reasoning. That doesn't withhold my curiousity from sniffing at 'complex math' (little bit of pun intended ) though.