Physics Computer Simulation: Optimum Angle for Drag Projectile

AI Thread Summary
The discussion centers on the implementation of a physics simulation for children, specifically focusing on projectile motion and the effects of drag. It is established that while 45 degrees is typically the optimal angle for maximum range in a vacuum, the presence of drag alters this angle, making it lower, such as 35 degrees in this case. The change in the optimum angle is attributed to the drag coefficient, which varies the relationship between the forces acting on the projectile. As the drag coefficient approaches zero, the optimal angle approaches 45 degrees again. The conversation reassures the original poster that their findings are correct and not indicative of a coding error.
Winchman
Messages
2
Reaction score
0
Hi everyone, I am hoping that someone may be able to advise me on an issue i currently have.
I am trying to implement a Physics computer simulation (only basic) designed for children learning about forces. One of the sub games that i have created is to fire a character from a cannon and find the optimum angle, which ends up at 45 degrees.
However, the second sub game models drag force on the projectile, taking into account its mass, area, drag coefficient, and the air density. The equations are solved using an RK4 method. Excuse my lack of knowledge on this subject, but I am not that hot on physics, i just have an interest which is why i decided to model this simulation as my final university project as a computing student. Now, the question i have is that I've always understood that 45 degrees is the optimum angle to travel the farthest, but when simulating the drag projectile, the optimum angle is now 35. Is this correct, in that 45 degrees would no longer be optimum, or am i going wrong somewhere?
Many thanks
 
Physics news on Phys.org
the optimum angle changes because of the drag coefficient and it is continuous. Meaning that the optimum angle depends on the value of the coefficient, and is not 35 for all drag coefficients, just that particular value. It approaches 45 as the coefficient goes to zero. I believe you can solve for it by minimizing the amount of work done by drag.
 
Yes, the angle will be less than 45 degrees.

An intuitive way to see this would be to realize that there is a force along the horizontal in the opposite direction to the velocity of the projectile. Add this force to the downward gravitational force and the resultant force acting on the projectile will be at a nonzero angle to the vertical. Hence, your angle of projection to maximize the range must be lower than 45 degrees. It would be similar (though not the same) as firing a projectile on an uphill slope. but you are now viewing the whole thing from a rotated point of view.
 
Last edited:
Ah, many thanks for your responses. Id suddenly became very worried when i realized that 45 was no longer optimum, and given my lack of knowledge id felt id made a mistake somewhere in the code. Again, your responses are much appreciated.
 
Thread ''splain this hydrostatic paradox in tiny words'
This is (ostensibly) not a trick shot or video*. The scale was balanced before any blue water was added. 550mL of blue water was added to the left side. only 60mL of water needed to be added to the right side to re-balance the scale. Apparently, the scale will balance when the height of the two columns is equal. The left side of the scale only feels the weight of the column above the lower "tail" of the funnel (i.e. 60mL). So where does the weight of the remaining (550-60=) 490mL go...
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Back
Top