You did not answer my questions.
Actually, I did, but you keep going tangent to the original commission of" for testing lift powders and tuning shell heights.
Here is one for you, by the way this one is actually useful for pyrotechnics for testing lift powders and tuning shell heights.
Given Shell diameter, mass, time of flight from mortar and back to ground (assume it is a dummy). Develop a series of equations to determine apogee, time to apogee, muzzle velocity and impact velocity. Also propose a way of solving the system. http://www.pyrosocie...ign/#entry83534
When testing 8 grams (IIRC) of BP from various sources we saw flight times of a cricket ball vary from a few seconds to 20 seconds! (muzzle velocities varying between 30m/s to over 135m/s and heights between 30m and over 300m). So unfortunately there is no general equation for black powder power that suits this purpose.
There are only really two things that can be measured with any sort or ease that allow the equations to be solved for the real world. These are muzzle velocity and time of flight. Clearly time of flight is the easiest to measure as just about everyone has a stopwatch on their phone. It is also possible to measure muzzle velocity relatively easily with the aid of a moderately high speed video recorder, but this is a great deal more effort (it does make the solution of the equations far easier however). http://www.pyrosocie...ge-2#entry83648
As I have already stated that: The XLS file I'm working on, will have 3 calculation tabs.
Internal Ballistics: calculates, newtons of force, acceleration, shell/mortar friction, & more.
External Ballistics: calculates, Reynolds number vs drag coefficient, Vertical Trajectory Calculations, & more.
Ballistics Data: all relevant ballistics data.
And I intend to include a, "Ballistic Chronograph" tab also.
Since your tests, with a cricket ball resulted in, "muzzle velocities varying between 30m/s to over 135m/s and heights between 30m and over 300m" I have decided to break this into Internal Ballistics and External Ballistics.
If you are measuring muzzle velocity why do you care what happens in the mortar?
By the way your equation only shows an average acceleration, this is not what happens inside a mortar, so why bother calculating something that does not actually happen?
As in (2) with the cricket ball, what is causing the varying muzzle velocities? Is it happening with the Internal Ballistics within the mortar, or the External Ballistics?
You talk about errors, clearly you don't fully understand the significance of errors. If you are half a second out on flight time with a stop watch (easily attainable by the way), then so what.
What does this mean in the real world with a parabolic function?
Well it means that a couple of meters on the apogee calculation? So less than a 1-2% on the final answer! so perfectly accurate enough for the purpose.
There will be a far bigger error in apogee calculation based on the drag correlation alone.
It has become equally clear to me, that you don't.
In the case of the calculation you have done in your last sheet many times the error in fact due to the constant drag taken for the calculation.
As I already have stated: External Ballistics: calculates, Reynolds number vs drag coefficient, Vertical Trajectory Calculations, & more. See attached picture.
What is the relevance of the paper on shell failures (prevention of muzzle break)?
Simply utilising commercial shell data, "Approximately 2.68 inches (68 mm) in diameter and the shells had an average total mass of approximately 4.8 ounces (130 grams)", to calculate W = mg = 0.13kg x 9.8 m/s2 = 1.274N
If you choose to "take it out of context", for whatever reason, WELL?
By the way the friction force is dependent on inclination, hence you can't say the frictional force of a horizontal tube equates to that of a vertical one.
And you thing I don't know that? I'm interested in measuring the total "frictional force" (W = mg = 0.13kg x 9.8 m/s2 = 1.274N) at horizontal as the maximum. As the mortar incline moves from horizontal to vertical the frictional force changes proportional to angle.
Actually, Ff = uN
Ff = frictional force
u = friction coefficients
N = Newtons (W = mg)