Anyone good at maths?

Anyone good at maths?

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FlavaDave

Original Poster:

213 posts

166 months

Thursday 7th November 2013
quotequote all
I've spent the past week or so trying to figure this out (see link)



It's a graph that plots the parabola of a projectile, affected by gravity and drag. You'd set this by inputting only two variables: Initial velocity and a landing point.

Once set, I'd like to input x values and read out the y value.

It's proving quite tricky. Anyone have any ideas?

It's not for any commercial gain by the way, just an interesting side project... :-)

thanks


tannedstamina

510 posts

136 months

Thursday 7th November 2013
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Not planning on launching some mortars at the Houses of Parliament or anything are you...?

xRIEx

8,180 posts

155 months

Thursday 7th November 2013
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This any use, or are you past this already? http://www.dummies.com/how-to/content/how-to-graph...

outofstepuk

1,242 posts

159 months

Thursday 7th November 2013
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I assumed this was going to be to help justify a car purchase. I'm good at those maths.

This, I have no assistance for you. Sorry.

FlavaDave

Original Poster:

213 posts

166 months

Thursday 7th November 2013
quotequote all
Haha - good at Man Maths. We can all do that...!

Yeah probably not what people are expecting when they click through.

See what happens

randlemarcus

13,599 posts

238 months

Thursday 7th November 2013
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Zad

12,762 posts

243 months

Thursday 7th November 2013
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Rather than producing some fancy calculus derived polynomial, I would iterate it. Start with a fixed mass, velocity and direction. Resolve the X and Y vectors. Add a fixed gravity (9.81m/s/s) to each iteration of the -Y speed scalar. Subtract drag from each component speed (proportional to the square of the component speed). Update the position for each component from the x&Y speed * time interval. Iterate through it at, say, 1 millisecond intervals until Y<=0 or whatever termination you choose.

TheEnd

15,370 posts

195 months

Thursday 7th November 2013
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What is X and Y supposed to be in those pictures?

FlavaDave

Original Poster:

213 posts

166 months

Thursday 7th November 2013
quotequote all
Zad said:
Rather than producing some fancy calculus derived polynomial, I would iterate it. Start with a fixed mass, velocity and direction. Resolve the X and Y vectors. Add a fixed gravity (9.81m/s/s) to each iteration of the -Y speed scalar. Subtract drag from each component speed (proportional to the square of the component speed). Update the position for each component from the x&Y speed * time interval. Iterate through it at, say, 1 millisecond intervals until Y<=0 or whatever termination you choose.
I think I understand what you're saying. If so it's a good idea to do ONE graph, but I need the formula so I can run it like a single function calculator for other inputs of Vo and X2.

FlavaDave

Original Poster:

213 posts

166 months

Thursday 7th November 2013
quotequote all
TheEnd said:
What is X and Y supposed to be in those pictures?
Distance and height in meters

PugwasHDJ80

7,558 posts

228 months

Thursday 7th November 2013
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Is the parabola a "normal" curve

if so then you can use pretty basic maths (UVAST)

FlavaDave

Original Poster:

213 posts

166 months

Thursday 7th November 2013
quotequote all
PugwasHDJ80 said:
Is the parabola a "normal" curve

if so then you can use pretty basic maths (UVAST)
No it depletes with "drag" and "gravity"

Zad

12,762 posts

243 months

Thursday 7th November 2013
quotequote all
FlavaDave said:
I think I understand what you're saying. If so it's a good idea to do ONE graph, but I need the formula so I can run it like a single function calculator for other inputs of Vo and X2.
Yes. You split (resolve) the polar coordinates (speed and angle) into two separate X and Y parts. You then modify the X and Y parts to take into account gravity and friction. Do this lots of times til it hits the ground, or the X point you are looking for. If you like, you can then convert the final X and Y back into polar (speed and angle) terms.

But, as you need a non-iterated version...

Google for trajectory with drag equation. It gets complicated very quickly. Interesting that the NASA K12 (i.e. for small children) goes into calculus.




Toltec

7,167 posts

230 months

Thursday 7th November 2013
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One of the problems is that there may not be a solution for a given velocity and landing point. You could use angle and x axis distance though.

MiseryStreak

2,929 posts

214 months

Friday 8th November 2013
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The height y of the projectile at distance x is given by



g: the gravitational acceleration—usually taken to be 9.81 m/s2 near the Earth's surface
θ: the angle at which the projectile is launched
v: the velocity at which the projectile is launched
y0: the initial height of the projectile
d: the total horizontal distance traveled by the projectile


Everything you need is in here:
http://en.wikipedia.org/wiki/Trajectory_of_a_proje...

ewenm

28,506 posts

252 months

Friday 8th November 2013
quotequote all
MiseryStreak said:
The height y of the projectile at distance x is given by



g: the gravitational acceleration—usually taken to be 9.81 m/s2 near the Earth's surface
θ: the angle at which the projectile is launched
v: the velocity at which the projectile is launched
y0: the initial height of the projectile
d: the total horizontal distance traveled by the projectile


Everything you need is in here:
http://en.wikipedia.org/wiki/Trajectory_of_a_proje...
If you ignore drag. The more involved maths at the bottom of the Wiki link include drag and shows how much that complicates matters! hehe

moreflaps

746 posts

162 months

Sunday 10th November 2013
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ewenm said:
MiseryStreak said:
The height y of the projectile at distance x is given by



g: the gravitational acceleration—usually taken to be 9.81 m/s2 near the Earth's surface
θ: the angle at which the projectile is launched
v: the velocity at which the projectile is launched
y0: the initial height of the projectile
d: the total horizontal distance traveled by the projectile


Everything you need is in here:
http://en.wikipedia.org/wiki/Trajectory_of_a_proje...
If you ignore drag. The more involved maths at the bottom of the Wiki link include drag and shows how much that complicates matters! hehe
Especially for trans-sonic projectiles....

FlavaDave

Original Poster:

213 posts

166 months

Monday 11th November 2013
quotequote all
ewenm said:
MiseryStreak said:
The height y of the projectile at distance x is given by



g: the gravitational acceleration—usually taken to be 9.81 m/s2 near the Earth's surface
θ: the angle at which the projectile is launched
v: the velocity at which the projectile is launched
y0: the initial height of the projectile
d: the total horizontal distance traveled by the projectile


Everything you need is in here:
http://en.wikipedia.org/wiki/Trajectory_of_a_proje...
If you ignore drag. The more involved maths at the bottom of the Wiki link include drag and shows how much that complicates matters! hehe
Yes, if you ignore drag AND know the angle of elevation of the muzzle.

Anyway. We've worked it out over the weekend. Thanks for all your input. :-)



MiseryStreak

2,929 posts

214 months

Monday 11th November 2013
quotequote all
FlavaDave said:
Yes, if you ignore drag AND know the angle of elevation of the muzzle.

Anyway. We've worked it out over the weekend. Thanks for all your input. :-)
I assumed you knew this as you said "You'd set this by inputting only two variables: Initial velocity and a landing point."
You must be entering a launch angle to plot the graph?

FlavaDave

Original Poster:

213 posts

166 months

Monday 11th November 2013
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No you can plot it without the angle. We've only just figured that out.