Finding rocket's distance at various angles after launch

[JDC]

This helps...

http://www.mrpk.org/Portals/11/usersdat ... 0TABLE.pdf

[JDC]

Does this sound right so far?

tan(45)=1
150*1=150

tan(10) = .176327
150*.176327 = 26.44905

150-26.44905 = 123.551

am I even close?

[buddy]

This helps...

http://www.mrpk.org/Portals/11/usersdat ... 0TABLE.pdf

why not use a calculator or a spreadsheet instead of a trig table?

tan(45)=1
150*1=150

what is the 150 coming from? are you doing something different than what you did at the starting?

tan(10) = .176327
150*.176327 = 26.44905
150-26.44905 = 123.551
am I even close?

i dont understand this at all. i showed you the regular trig way to do this:

Code: Select all

rocket path: l *. v a / | ` . i u / | ` . e n / |h ` . w c /45* | 10* ` . e h *---------*------------------* r |<===========150============>|

Since the triangle on the launch side is 45 degrees then the base is h too. Then the base of the viewer's triangle is 150-h.

so do like i said:

Code: Select all

triangle for A: l A*. v a / | ` . i u / | ` . e n / |h ` . w c /45* | 10* ` . e h *---------*------------------* r | h | 150-h | |<===========150============>|

So do the tangent of the 10 degrees & make it equal to h/(150-h).

so thats tan(10*)=h/(150-h)
(i don't know what 150*.176327 = 26.44905 is?)
solve tan(10*)=h/(150-h) for h.

[JDC]

Ok, when I use excel for TAN(10) I get 0.176327

So, if you're saying that "h/(150-h)" needs to equal the same then I'm not sure how to formulate that.

I don't know the formula, but, by deduction it's about "22.4845"

How do I get 22.5?

[JDC]

Ok, I figured it out finally!

I just created a SpreadSheet and found all the formulas 1 at a time...

Thanks for helping, I think you might have left out a critical piece re:SinA, once I read about that it all made sense

In the spreadsheet, it only requires A, B and c

[nona.m.nona]

No information was omitted from the step-by-step solution method provided to you earlier. Rather, you elected to solve in a completely different manner. As a result, you used different steps.

Thank you for posting your solution method for the benefit of other viewers. As is so often the case, your exercise admitted of more than one valid solution method.

[JDC]

Is there a quicker way of doing this?

Sorry, but, I could not understand the solution you tried to walk me through, would you mind posting something similar?

If there is an easier way I'd like to know, the above steps seemed to be a little overkill, but, at least I was able to get the answer.

I did translate this into C# code if anyone is interested....

public class Collision
{
public double X1; //Left Side
public double X2; //Right Side
}

public Collision GetCollisionPoint(int baseWidth, double angleA, double angleB)
{
int c = baseWidth;
double sinA = Sin(angleA);
double sinB = Sin(angleB);
double sinC = Sin(180 - angleA - angleB);
double tanB = Tan(angleB);
double a = c*sinA/sinC;
double height = a*sinB;
double length = height/tanB;

return new Collision {X1 = baseWidth - length, X2 = length};
}

public double Sin(double value)
{
return Math.Sin(Degrees(value));
}

public double ArcTan(double value)
{
return Math.Atan(Degrees(value));
}

public double Tan(double value)
{
return Math.Tan(Degrees(value));
}

private double Degrees(double value)
{
return value * Math.PI / 180;
}

[JDC]

Is there anyway to post zip files, I'd be happy to share the Excel SpreadSheet. If anyone else runs into this questions I'm sure it would be useful.

Any suggestions for uploading files?

[JDC]

The Spreadhseets are here.

There is just 1, but, 3 different formats in case someone doesn't have Excel. There is open document format as well, Open Office is free.

Hope someone finds this useful.

https://drive.google.com/file/d/0B1m00D ... sp=sharing

[nona.m.nona]

Is there a quicker way of doing this?

Of doing which?

Sorry, but, I could not understand the solution you tried to walk me through, would you mind posting something similar?

Certainly. It is useful to begin with the coded "picture" as provided by the other poster:

Code: Select all

triangle for A: l A*. v a / | ` . i u / | ` . e n / |h ` . w c /45* | 10* ` . e h *---------*------------------* r | h | 150-h | |<===========150============>|

For a similar set-up, suppose that the distance between the "launch" location and the "viewer" location is "d", and that the angles are "L" and "V", respectively; let the distances be "x" and "y", respectively.

Code: Select all

triangle for A: l A*. v a / | ` . i u / | ` . e n / |h ` . w c / L* | V* ` . e h *---------*------------------* r | x | y | |<===========d==============>|

Then tan(L*) = h/x, so h = x tan(L*); tan(V*) = h/y, so h = y tan(V*). Noting that y = d - x, one obtains h = (d - x) tan(V*). Setting the two expressions for "h" equal (as h must necessarily equal itself), one obtains (d - x) tan(V*) = x tan(L*):









As all values other than x are known, the solution follows.