Trigonometry Formulas - Page 2

Pisces · 09-11-08, 06:37 PM

@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 · 09-12-08, 08:17 AM

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 · 09-12-08, 09:02 AM

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 · 09-12-08, 12:02 PM

Quote:

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.

ReallyDedPoet · 09-12-08, 12:42 PM

Quote:

Originally Posted by Redwine

Look at my signature, there is a complete procedure to manually shoot... i made long time a go for a SH II Mod... but it is valid for any subsim, maths are the same.

Nice guide RW

Will have to give it a go at some point as I have never used Manual Targeting with SH3, but love it with 4.

RDP

JMV · 09-13-08, 07:57 AM

@ Pisces

aeronautical engineer hmmmm ? I shouldn't be too far off the mark if I say you must be an FSX player... That's my best guess. If not, try it, it's a great sim too.

JMV ( and tks again )

Pisces · 09-13-08, 01:01 PM

Yeah, I know about FS(X). Have played the earlier versions, but not FSX yet. In college we even had a real cockpit section of a Fokker F28 (pilot side and a partial co-pilot side) built into a stationary simulator running FS98. Hands on the yoke and feet on the pedals. That was big fun.

Hmm, found a picture of it on a site of a buddy of mine.

However, that education wasn't finnished as I planned it. And with the years passing my interest in aviation took a bit of a nosedive.... and I ended up here in the water. :rotfl:

JMV · 09-13-08, 04:34 PM

@Pisces,

You should seriously consider buying FSX Deluxe Edition. It's a great improvement from the previous versions... and in conjunction with SH3, you can be pretty sure, to be glued to you PC 24/7:rotfl: . Got 224 different planes in total, not including the different versions of every one of them... And favorites folder on FSX is... well, 401 Mb, a bit smaller than SH3 though. JMV

09-11-08, 06:37 PM	#16
Pisces Silent Hunter Join Date: Dec 2004 Location: AN9771 Posts: 4,892 Downloads: 300 Uploads: 0	@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. Last edited by Pisces; 09-11-08 at 06:59 PM.

09-12-08, 08:17 AM	#17
JMV Sonar Guy Join Date: Jun 2008 Location: Galway-IRELAND Posts: 397 Downloads: 179 Uploads: 0	Trigonometry Formulas 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

09-12-08, 09:02 AM	#18
JMV Sonar Guy Join Date: Jun 2008 Location: Galway-IRELAND Posts: 397 Downloads: 179 Uploads: 0	Trigonometry Formulas 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

09-13-08, 07:57 AM	#21
JMV Sonar Guy Join Date: Jun 2008 Location: Galway-IRELAND Posts: 397 Downloads: 179 Uploads: 0	Trigonometry Formulas @ Pisces aeronautical engineer hmmmm ? I shouldn't be too far off the mark if I say you must be an FSX player... That's my best guess. If not, try it, it's a great sim too. JMV ( and tks again )

09-13-08, 04:34 PM	#23
JMV Sonar Guy Join Date: Jun 2008 Location: Galway-IRELAND Posts: 397 Downloads: 179 Uploads: 0	Trigonometry Formulas @Pisces, You should seriously consider buying FSX Deluxe Edition. It's a great improvement from the previous versions... and in conjunction with SH3, you can be pretty sure, to be glued to you PC 24/7:rotfl: . Got 224 different planes in total, not including the different versions of every one of them... And favorites folder on FSX is... well, 401 Mb, a bit smaller than SH3 though. JMV

09-13-08, 01:01 PM	#22
Pisces Silent Hunter Join Date: Dec 2004 Location: AN9771 Posts: 4,892 Downloads: 300 Uploads: 0	Yeah, I know about FS(X). Have played the earlier versions, but not FSX yet. In college we even had a real cockpit section of a Fokker F28 (pilot side and a partial co-pilot side) built into a stationary simulator running FS98. Hands on the yoke and feet on the pedals. That was big fun. Hmm, found a picture of it on a site of a buddy of mine. However, that education wasn't finnished as I planned it. And with the years passing my interest in aviation took a bit of a nosedive.... and I ended up here in the water. :rotfl: