Projectile angle that considers canon length

In summary: We can solve for ##y_0## by solving for ##y## at ##(x_0,y_0)##. Now we know that ##y_0## is the height of the ball, ##y## is the distance to the cup, and ##V_0## is the velocity of the ball at the cannon.
  • #1
Edvin
5
2
Greetings,

Most projectile problems are presented from the muzzle of a cannon; however, how do we solve the problem if the velocity is known at the muzzle; but the angle is controlled at the base of the canon.
The issue is that, when the canon angle changes, so does the muzzle location!

The problem that I am trying to solve is for a small robot that shoots a ball into a coffee cup. I can control the angle; but, not the velocity. When I try to solve the problem, I end-up with a fourth degree polynomial.
I'm not concerned with air drag; because, it is negligible (at least in the simulation I created in algodoo.

Here is how I approach the problem:
h = v0 t + 1/2 g t2
given that vx = v cosθ and vy = vsinθ
Also, distance traveled is given by d = vx t; therefore, d= t v cosθ, where t = d/(v cosθ)
Next, I add the extra height and distance as the canon angle changes.
The extra horizontal length for given barrel length L is going to be L - L cosθ.
So, t = (d +L - L cos θ)/ (v cos θ)
The extra height is L sin θ
So, the first equation changes to:
h = L sinθ + vy t + 1/2 g t2
h = L sinθ + v sinθ [ (d +L - L cos θ )/ (v cos θ) ] + 1/2 [(d +L - L cos θ )/ (v cos θ)]2
...
2v2h2 - 2L2 - 2v2(d+L)tanθ - g(d+L)2 - g(d+L)2 tan2θ = 2 L g sqrt(1+tan2θ)
...
x1 tan4θ + x2 tan3θ + x3 tan2θ + x4 tanθ + x5 = 0​

If this is right, I would have to write an algorithm to solve a 4th degree polynomial. Alternatively, I can ignore the canon length, and then loop through different angles until I calculate the right value.
However, I would prefer to solve the problem down to an equation and then plug it in.

Please let me know if I'm mistaking or if there is another way of solving this problem.

Tnx,
Edvin
 
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  • #2
I would do it like this:
First, I assume that you only know the velocity in the direction of the cannon, V_0, at the exit of the barrel. I call the horizontal distance x, and the vertical distance y. Gravity is pointing down, in the negative y-direction.
The x-component of velocity is [itex]V_0 \cos (\theta)[/itex] and the y-component is [itex]V_0 \sin (\theta)[/itex].
The exit of the barrel is at a height of [itex]y=L \sin (\theta)[/itex].

Now we call the exit of the barrel P=(x1,y1)=(0,0). The next time the ball is at y= 0 is given by the equation for y:
[itex]y=V_0 \sin (\theta)t - \frac{1}{2}gt^2=0[/itex]

The total distance traveled can be obtained from the equation for x. The 'real' ground is [itex]L \sin (\theta)[/itex] lower.
Now you need to calculate the time it takes the ball to fall down this distance, and then calculate the extra x-distance.
The angle and velocity components at (x2,y2) are known from the symmetry of the parabola.
 
  • #3
Tnx for looking at this...

You are right, I can determine V_0 at the exit with a few experiments.
I also know the distance to target (cup) from base of my canon (determined with sensors).
I certainly understand what you mean by the symmetry of velocity and angle at (x2,y2).

I'm not sure how to drive to θ from the known variables.
I'll try to work the problem again, see where I end-up.

BTW, how do you add the equations in this forum?
EDIT: I found, this how to link
 
Last edited:
  • #4
I tried the problem a few more times; but, ultimately I ended up with a 4th degree polynomial.
I reached out to a physics professor and he was kind enough to share that a general solution would be very complicated and that simplifying via approximation would give a reasonable result.

Recall that my goal is to shoot an object into a cup.
Since the farthest distance for a projectile is a 45 degree angle; we know that my angle must be 45 degrees and higher.
The closer the target (cup) is to the cannon, the less significant is the cannon length.
We will approximate the projectile origin at the top of the cannon to be ##\mathcal{L}## (as appose to ##\mathcal{L}\ sin \theta##) and that we'll approximate the horizontal offset from the base of the cannon (since ##\mathcal{L}\ cos \theta## is almost zero). [Note ##\mathcal{L}## is our cannon length]

In my original post I showed t = (d +L - L cos θ)/ (v cos θ).
So, we'll revise it such that distance ##\mathcal{d}## is from the base of the cannon to the cup. And we'll endup with ## t = \frac {d- l\ cos \theta} {v\ cos \theta} \approx \frac{d}{v\ cos \theta}##
We'll also change our initial y offset to cannon length since ##\mathcal{L}\ sin \theta \approx \mathcal{L}##. Now we end up with
##y = y_0 + V_0\ t - \frac{1}{2}\ g\ t^2##
##y = L + V\ sin\theta\ ( \frac{d}{v\ cos \theta} ) - \frac{1}{2}\ g\ (\frac{d}{v\ cos \theta})^2##

And that leads to:
##\frac{gd^2}{2v^2}tan^2\theta-d\ tan \theta+(\frac{gd^2}{2v^2}+y-l)=0##

And now we have our quadratic equation:
##tan \theta = \frac{v^2+\sqrt{2v^2\ g(l-y)+v^4-d^2g^2}}{dg}##

Incidentally, if anyone wants to try 4th degree polynomial; I found the following threads:
4th degree quadratic
general formula for 4th degree
Yet another 4th degree tutorial
 
  • #5


Hello Edvin,

Thank you for sharing your approach to solving this problem. It seems like you have a good understanding of the physics involved and have come up with a valid approach. However, I would suggest considering the use of trigonometric identities to simplify your equations. For example, you can use the identity cos^2θ + sin^2θ = 1 to eliminate the square terms in your equation. This may make it easier to solve and avoid the use of a 4th degree polynomial.

Another approach could be to use vector mathematics to solve the problem. By representing the velocity and angle as vectors, you can use vector addition and trigonometric functions to calculate the trajectory of the projectile. This may also be a more efficient and accurate method.

I hope this helps and good luck with your project!

Best,
 

1. How does cannon length affect the projectile angle?

The length of a cannon can affect the angle at which the projectile is launched. A longer cannon will typically have a higher angle of elevation, resulting in a longer range for the projectile. This is due to the longer barrel providing more time for the projectile to accelerate before leaving the cannon.

2. Is there an optimal angle for a specific cannon length?

The optimal angle for a specific cannon length depends on various factors such as the type of projectile, launch velocity, and desired range. However, in general, a longer cannon will have a higher optimal angle of elevation compared to a shorter cannon.

3. How does the weight of the cannon impact the projectile angle?

The weight of the cannon can impact the projectile angle by affecting the recoil force. A heavier cannon will typically have a stronger recoil force, resulting in a higher angle of elevation for the projectile. This is because the heavier cannon will absorb less of the recoil energy, allowing more of it to be transferred to the projectile.

4. Can the projectile angle be adjusted by changing the cannon length?

Yes, the projectile angle can be adjusted by changing the cannon length. A longer cannon will result in a higher angle of elevation, while a shorter cannon will result in a lower angle of elevation. This can be useful for adjusting the range or trajectory of the projectile.

5. How does the wind affect the projectile angle for a specific cannon length?

The wind can have a significant impact on the projectile angle for a specific cannon length. Strong winds can cause the projectile to deviate from its intended path, resulting in a different angle of elevation. This is why it is important for cannon operators to take wind speed and direction into account when determining the optimal angle for their cannon.

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