Staredit Network > Forums > Lite Discussion > Topic: Find theta
Find theta
Oct 16 2010, 4:21 pm
By: Aristocrat  

Oct 16 2010, 4:21 pm Aristocrat Post #1





I so confuzed :S. Arrived at this equation after I'm halfway done with a problem and can't figure out a way to isolate theta. The teacher insists that it's possible but will not surrender the solution. Anyone bright enough to see a simple way to get the answer?

(Δx, Δy, and v are constants, but that shouldn't really matter much. )



None.

Oct 16 2010, 4:54 pm CecilSunkure Post #2



Try #1:
0 = tan(θ)Δx - Δy + (4.9(Δx2 / (vcos(θ))2))

0 = tan(θ)Δx - Δy + (4.9(Δx / (vcos(θ)))) · (4.9(Δx / (vcos(θ))))


Try #2:
0 = tan(θ)Δx - Δy + (4.9(Δx2 / (vcos(θ))2))

0 = tan(θ)Δx - Δy + (4.9(Δx2 · (vcos(θ))-2))

-(4.9(Δx2 · (vcos(θ))-2)) = tan(θ)Δx - Δy

(Δx2 · (vcos(θ))-2) = -4.9(tan(θ)Δx - Δy)


I don't see a straight forward way to do this, since you have two separate theta variables. Since there are two variables, every time you try to isolate one, you modify the other, and thus have to do the opposite of your modification to isolate the first one, putting you back to where you started. You can't isolate theta unless you find a single set of formula modifications that frees up both variables at once, and I either can't see how, or it's impossible.



None.

Oct 16 2010, 7:23 pm payne Post #3

:payne:

Not too sure, but I can try to help A BIT:

Tan = Sin/Cos.
Like this, you get 2 Cos at the bottom of fractions.
And with Delta Y, multiply by Cos/Cos.
From there, try to isolate some of the cos in front of the whole equation.
Once that's done, you should get something like:

0 = cos(θ) * (BLAH BLAH BLAH)

All you need to do is to divide 0 by "BLAH BLAH BLAH", and you're left with cos(θ) = 0.
cos-1(θ)=cos-1(0).

But I'm really not sure and I haven't calculated anything. It was just what I had in mind. ^^



None.

Oct 16 2010, 7:34 pm Aristocrat Post #4



Quote from payne
Not too sure, but I can try to help A BIT:

Tan = Sin/Cos.
Like this, you get 2 Cos at the bottom of fractions.
And with Delta Y, multiply by Cos/Cos.
From there, try to isolate some of the cos in front of the whole equation.
Once that's done, you should get something like:

0 = cos(θ) * (BLAH BLAH BLAH)

All you need to do is to divide 0 by "BLAH BLAH BLAH", and you're left with cos(θ) = 0.
cos-1(θ)=cos-1(0).

But I'm really not sure and I haven't calculated anything. It was just what I had in mind. ^^





None.

Oct 16 2010, 8:10 pm payne Post #5

:payne:

Quote from Aristocrat
Quote from payne
Not too sure, but I can try to help A BIT:

Tan = Sin/Cos.
Like this, you get 2 Cos at the bottom of fractions.
And with Delta Y, multiply by Cos/Cos.
From there, try to isolate some of the cos in front of the whole equation.
Once that's done, you should get something like:

0 = cos(θ) * (BLAH BLAH BLAH)

All you need to do is to divide 0 by "BLAH BLAH BLAH", and you're left with cos(θ) = 0.
cos-1(θ)=cos-1(0).

But I'm really not sure and I haven't calculated anything. It was just what I had in mind. ^^

I've made 0/something, and not something/0. I can't see the relevancy of your video. :/



None.

Oct 16 2010, 8:13 pm CecilSunkure Post #6



0/something = 0

Just making note.



None.

Oct 16 2010, 8:18 pm Vrael Post #7



Is tan(theta) = delta(y)/delta(x)?



None.

Oct 16 2010, 8:44 pm payne Post #8

:payne:

Quote from CecilSunkure
0/something = 0

Just making note.
Hence my quick deduction that Theta was equal to cos-1(0)...
I thought the 0 was a good thing we could use to our advantage to nullify a lot of things with divisions or multiplications. ;o



None.

Oct 16 2010, 9:13 pm Aristocrat Post #9



Quote from CecilSunkure
0/something = 0

Just making note.
cos(theta) was in the denominator, so if payne arrived at the conclusion that cos(theta) = 0, then the original expression has divisions by zero, which is clearly not the case.

Quote from Vrael
Is tan(theta) = delta(y)/delta(x)?
I wish lol. Here's a picture: http://img840.imageshack.us/img840/5303/tempao.png

(doesn't show well on black background)



None.

Oct 16 2010, 9:31 pm Vrael Post #10



Are you allowed to make up your own equations for this, or is it just an excersize in algebra to solve?



None.

Oct 16 2010, 9:45 pm DT_Battlekruser Post #11



So tan θ = dy/dx at the x-intercept? Is this v?



None.

Oct 16 2010, 9:50 pm Aristocrat Post #12



Quote from Vrael
Are you allowed to make up your own equations for this, or is it just an excersize in algebra to solve?
Sure, why not. I would like to know if it's possible to explicitly solve the equation in the OP for θ, however. :)

Quote from DT_Battlekruser
So tan θ = dy/dx at the x-intercept? Is this v?
Yes, this is v.

http://www.staredit.net/?p=shoutbox&view=1183

Basically: Given v, find θ for which a projectile launched from (0, 0) will pass through point N, with coordinates (Δx, Δy). Gravity is -9.8 m/s2 in the negative-y direction.



None.

Oct 16 2010, 9:52 pm Vrael Post #13



Oh I think I got it.
2 hints:
sec^2 = 1 + tan^2, and the quadratic formula.



None.

Oct 16 2010, 10:14 pm Aristocrat Post #14



Oh my.
Quote from Vrael
Oh I think I got it.
2 hints:
sec^2 = 1 + tan^2, and the quadratic formula.


O_o

I can probably simplify it a little bit more. Hmm...

EDIT> Okay, this is why the 400 pixel image limit is stupid. Can we please just get rid of it?



None.

Oct 16 2010, 10:48 pm Vrael Post #15



Okay, right idea, wrong substitution.

Instead of substituting sqrt(sec^2 - 1) for tan(theta), try substituting 1 + tan^2 for sec^2.
That way if tan(theta) = x, you have an equation of the form
0 = ax^2 + bx + c, instead of ax + bsqrt(x) + c
Should simplify things for you.



None.

Oct 16 2010, 11:22 pm Sacrieur Post #16

Still Napping

well vrael beat me to it... Demon contacted me asking to help =P, what a nice guy haha.

Okay, so you know all those constants? Fuck them, they're constants and they can fuck themselves. Basically they're just trying to overwhelm you. So yeah, simplify this way:

a = (4.9 delta(x)^2)/v^2
b = delta(x)
c = -delta(y) + a

and do what vrael said above and substitute sec^2(theta) with 1 + tan^2(theta). You will distribute the a to the 1 and to the tan^2(theta), then lump -delta(y) and a together and replace it with c (because they're both constants they're going to have butt sex with eachother =P).

then omigosh, you get...

tan(theta) = quadratic formula

theta = arctan(quadratic formula)

Plug in values where needed and get the answer =P



None.

Oct 16 2010, 11:29 pm Vrael Post #17



Quote
(because they're both constants they're going to have butt sex with eachother =P).
I'm sorry I must have missed this part of the class when I took basic algebra, can you explain this please?



None.

Oct 16 2010, 11:32 pm Sacrieur Post #18

Still Napping

Quote from Vrael
Quote
(because they're both constants they're going to have butt sex with eachother =P).
I'm sorry I must have missed this part of the class when I took basic algebra, can you explain this please?

xD, my school had budget cuts so they combined my algebra and sex ed classes >_>



None.

Oct 17 2010, 12:19 am Aristocrat Post #19



^ Post of the year.

Also, woot! Got a much nicer expression to work with.



I have a feeling that something might be off, though. Can't quite put my finger on it and I checked the work.



None.

Oct 19 2010, 2:36 am rockz Post #20

ᴄʜᴇᴇsᴇ ɪᴛ!

I get a quadratic equation:
btan2 + tan + (b-dy/dx) = 0 where b=4.9dx/v2

tan=(-1+-sqrt(1-4b(b-dy/dx)))/2b

looks right to me


oh god what the fuck.



"Parliamentary inquiry, Mr. Chairman - do we have to call the Gentleman a gentleman if he's not one?"

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