The Determination of the Launch Angle of a Projectile

[Kieran Hall]

TL;DR Summary

The determination of a projectile's launch angle from a boom of angle theta

Hi, I am new here to the forum and I am having trouble with a project that I am undertaking with some friends.
We are trying to build a firefighting robot.

I am trying to derive an expression to solve for the launch angle theta of the water so that at x (meters), the projectile will be at 0.33 (meters) of height.

The water will be launched from a boom arm of length L (meters) as opposed to being projected at an angle from a height of 'h' (meters).

Obviously as the boom angle is changed, the height that the projectile is launched from will change as well.

I have tried to incorporate this height of the boom into my equation as 'L.Sin(theta)'.

This is the difficulty that I am having as I don't know if my equation can be solved and if not, I'm unsure how to calculate the required boom angle.

Any help will be extremely appreciated.

Thanks,
Kieran

 

[berkeman]

Welcome to the PF.

I don't think that you can use the simple equations for projectile motion when doing calculations for a projected water stream. It seems like air resistance is very non-negligible for a water stream projected from a hose.

Have you searched online for treatments of this problem? There are certainly equations for projectile motion that include air resistance, but even those may not be accurate enough for a water stream...

 

[Kieran Hall]

Thank you,

Yes, we have considered the affect of air resistance on our stream of water however, we have decided to ignore it for the moment as the diameter of the nozzle is approximately 1mm (we're building a model robot) therefore, there would be very little resistance to our tiny stream of water.

We are more troubled over calculating the launch angle at the moment.

Thank you

Kieran

 

[berkeman]

I didn't go through your calculations past the first couple of lines, but it seems like your problem definition is over-constrained (unless I'm just not understanding the problem statement).

Do you want to *both* just clear the obstruction that is 0.33m high, *and* hit the spot that is 0.15m high and 0.9m past the obstruction? You can probably find an angle that clears the obstruction and hits that final spot, but to *just barely* clear the obstruction constrains the rest of the path of the projectile, it would seem...

 

[Kieran Hall]

Yes, the water stream has to clear the wall and hit the target.
For my 1st calculations I am purely concerned with determining the min and max angle for theta that will clear the wall.

The wall is actually 0.3m high and I had added on an additional 0.03m of height to the projectile's path to allow for clearance.

 

[berkeman]

But do you understand what I meant by "overconstrained"?

 

[Kieran Hall]

I'm sorry but I don't think I do

 

[berkeman]

It could be that I'm just understanding the problem statement. If the problem is to clear the wall and hit the target, that is straightforward (leaving air resistance aside). But if the problem is to just barely clear the wall and hit the target, that probably can't be done (although maybe you can adjust the water pressure / Vi value, but I'm not sure). Once you say that you are just barely going to clear the wall, that determines the rest of the projectile's flight path.

Is the intent to use the water pressure / projectile velocity as a variable that you can tune to try to hit both of those points (just over the top of the wall and the final target)?

 

[Kieran Hall]

Yes, the problem is just to clear the wall and hit the target.
I'm sorry, I may have been a little unclear.

No, the velocity is going to be fixed and it is purely the angle that will be used to hit target.

 

[Kieran Hall]

Yes, the problem is just to clear the wall and hit the target.
I'm sorry, I may have been a little unclear.

No, the velocity is going to be fixed and it is purely the angle that will be used to hit target.

The water does not have to just barely clear the wall, I was just assuming the wall to be 0.33m in my calculations so that there is 3cm of minimum clearance, no matter what the theta angle is.

 

[berkeman]

I'd recommend working through the equations to see how you do not have enough control over the projectile motion to do both of those things at the same time. I'll look more at the work you've done so far to see if I can offer any suggestions. I was confused by this part, though, are those symbols integral signs?

 

[Kieran Hall]

Thank you so much,

No they are just dashes ( / ), breaking up my initial constraints.
What I meant by them was that Sy = (h + L.SinTHETA) and its also = 0.33m
I'm sorry, I should have been clearer.

 

[berkeman]

That's okay. Is x a variable that you can tune to try to hit both points (top of wall and the target)?

BTW, it's probably a good time to learn LaTeX for posting math equations online and in your homework papers. There is a good tutorial under INFO, Help at the top of the page. It makes what you type into the forums much easier to read and comment on.

 

[Kieran Hall]

How we are designing the robot to function is that the robot will be able to determine, if stopped in its current position, if there is an angle theta where it is possible to clear the wall and hit the target.
If the robot determines no angles possible for these conditions, then it will move forward an arbitrary distance to a new x position and recalculate.

So yes, the distance x is a variable that can be changed so that the water stream can hit the target.

My idea for determining the angle theta was to perform three calculations.
1) Determine a max and min angle for theta to clear the wall with a minimum of 0.03m of clearance.
2) Determine an angle for theta that will hit the target.
3) Compare the angles and if the angle determined in step '2' is within the range calculated in step '1', then the robot can fire at the angle determined in step '2'.

These calculations would be very straight forward however, I am getting hung up on how to take into account the change in height of the boom arm (L) as the angle of theta is changed.

Thanks

 

[Kieran Hall]

That's okay. Is x a variable that you can tune to try to hit both points (top of wall and the target)?

BTW, it's probably a good time to learn LaTeX for posting math equations online and in your homework papers. There is a good tutorial under INFO, Help at the top of the page. It makes what you type into the forums much easier to read and comment on.

Thank you for the suggestion, that is a very good idea.
I will definitely be learning this software!

 

[berkeman]

These calculations would be very straight forward however, I am getting hung up on how to take into account the change in height of the boom arm (L) as the angle of theta is changed.

Because the launch point changes a little as the angle theta changes? You just need to alter the x,y position of the launch point as you change the angle. I'd define the base of the launch arm as Xo,Yo, then figure out where the launch point is based on the arm's length and the angle. It looks like you were starting to try to do that in your initial equations, but just weren't real clear about separating the two things?

 

[berkeman]

So I wouldn't define x as the distance from the tip of the launching arm, instead I'd define it to the base of the launcher arm at $(x_0,y_0)$, and make the launching tip position $(x_l,y_l)$ something that is part of the calculations...

$$x_l = x_0 + Lcos(\theta)$$
$$y_l = y_0 + Lsin(\theta)$$

 

[berkeman]

BTW, you can see the LaTeX that I used in my last post by clicking on "Reply" in the lower right of my reply. That will insert my source in the Edit window, showing the LaTeX that I used.

Note that in-line LaTeX is inserted with [ itex ] and [ /itex ] tags (without the spaces), and standalone LaTeX lines use [ tex ] [ /tex ] tags.

 

[Kieran Hall]

That's a good idea, I never thought about the fact that the arm would be moving in the x direction as theta is altered.

How would you suggest that I solve for theta with your approach?

 

[berkeman]

How would you suggest that I solve for theta with your approach?

Just write the equations and solve for $\theta$ and X_o ...

 

[PeroK]

So yes, the distance x is a variable that can be changed so that the water stream can hit the target.

My idea for determining the angle theta was to perform three calculations.
1) Determine a max and min angle for theta to clear the wall with a minimum of 0.03m of clearance.
2) Determine an angle for theta that will hit the target.
3) Compare the angles and if the angle determined in step '2' is within the range calculated in step '1', then the robot can fire at the angle determined in step '2'.

These calculations would be very straight forward however, I am getting hung up on how to take into account the change in height of the boom arm (L) as the angle of theta is changed.

Thanks

Why not calculate the angle(s) to hit the target? Pick the larger angle. Then check where the projectile is at a distance $x$. If it's not above the wall, then you can't hit the target.

Ultimately there will be a range for $x$ where the target can be hit. If you find this range then you are done.

The variation in launch position with angle is a complication, it's true. Maybe solve first without this, then add it?

 

[DEvens]

The larger launch angle will (probably) have more clearance room with the intermediate barrier. But it will have a longer flight time, and so have more time for the forward velocity to get reduced by air friction. So the larger launch angle is likely to be less accurate for distance.

A 1 mm droplet is likely to have substantial air resistance. Remember your cube-square ratio. A 1 mm droplet will have a relatively large area-to-volume ratio compared to a larger droplet.

Once you calculate the idealized path the water takes you should do some measurements. That will give you a correction factor to apply to distance. This correction factor is likely to depend on the launch angle.

You probably need to do some measurements anyway, since you need to determine the speed the water comes out of the nozzle. You can probably get that from things like the maximum height the water can reach. And you can back it up with measurements of the diameter of the nozzle and the flow rate of water through it.

Depending on how long the boom is, you might find that the speed the water comes out is angle dependent as well. Though that effect might well be very small. With the boom at a higher angle there will be a back-pressure due to the column of water. Measurements will put that issue to rest.