42 replies to this topic

[Crazy Cat]

Your equation does not balance, you have kg.m/s^2 on one side and m/s on the other so the = sign can't apply.

It's not in equation form, but more as an explanation. Let me rephrase it as, net force N (kg.m/s^2) acted upon the shell, at the mortar muzzle, THAT has accelerated the shell to a velocity (m/s) at the muzzle. Newtons. http://en.wikipedia....ki/Newton_(unit)

how do you intend to solve the force applied by the lifting charge?

Once an accurate muzzle velocity (m/s) is determined, by the ballistic chronograph, we can calculate the shell acceleration in the mortar, and force N (kg.m/s^2) on the shell to acceleration it.

The problem is the gap as you are aware.

Yes, indeed.

This is a tricky one, but not unsolvable. I guess you are going to do an annular gas flow calculation about a sphere, I guess Navier Stokes will offer the best modelling possibilities. But with a few assumptions it may be possible to simplify the solution somewhat without affecting accuracy too much.

Are you going to do a gas pressure iteration based upon standard burning rates, gas production and the usual thermodynamics including the adiabatic index.

I have not looked into the modelling in any detail, but I guess you will have to rely on published drag tables to solve the derived fluid flow equations in this instance.

I look forward to seeing your solution.

I'll look into this once accurate muzzle velocity (m/s) is determined, by the ballistic chronograph.

Edited by Crazy Cat, 21 November 2014 - 11:11 PM.

[digger]

OK

 

If you intend to measure the muzzle velocity, what is the point of calculating the acceleration force in the mortar?

 

As you know there are plenty of programs out there that can calculate the flight path once the muzzle velocity is known.

 

In fact knowing the muzzle velocity only allows you to calculate an average force directly, as we know this is not what happens in the mortar, so the model can't actually represent what happens inside the tube without doing all of the maths that I indicated in the last post. In which case you don't need to measure the muzzle velocity other than to check your maths.

 

So what useful information can be derived?

 

Still all seems rather long winded way of accomplishing something that can be done with a stopwatch and some simple maths, maybe I have missed the point somewhere.

 

Anyway, lets see how you get on with it.

[Crazy Cat]

If you intend to measure the muzzle velocity, what is the point of calculating the acceleration force in the mortar?

A "gun barrel" of length (l), and the needed velocity (Ve), the acceleration (a) is provided by the following formula:

As you know there are plenty of programs out there that can calculate the flight path once the muzzle velocity is known.

Yes, I'm certain there is, and I can post a site that has some free and purchased external ballistic software.
 

In fact knowing the muzzle velocity only allows you to calculate an average force directly, as we know this is not what happens in the mortar, so the model can't actually represent what happens inside the tube without doing all of the maths that I indicated in the last post. In which case you don't need to measure the muzzle velocity other than to check your maths.
 
So what useful information can be derived?

 
This will be addressed later on with the dummy shell, having electronic sensors to monitor "what happens in the mortar" an Accelerometer and ??

Still all seems rather long winded way of accomplishing something that can be done with a stopwatch and some simple maths, maybe I have missed the point somewhere.

There is ALWAYS the human error factor when taking measurements. That's why we have the, Absolute Error, Percentage and Relative Error, and The Combination of Errors.

With a ballistic chronograph, the muzzle velocity (m/s) can be accuracy measured within +/- 0.5-1%

[Crazy Cat]

I'll be shooting these babies out of a 2-1/2-inch HDPE mortar that I own. It's the smallest one that I have that these shells will fit into, and they will be a slightly loose fit. I weigh out 1/2-ounce of 2FA lift powder for each shell. Fg or FFg commercial sporting black powder could also be used. http://www.skylighte...ster-shells.asp

In the current tests, the shells were 3-inch (75-mm) Thunderbird brand “Color Peony-Gold” product number TBA-105. The shells were approximately 2.68 inches (68 mm) in diameter and of typical paper construction with two time fuses. On average, the shells had a total mass of approximately 4.8 ounces (130 grams), and with approximately 1.3 ounces (36 grams) of lift powder. The shells contained approximately 2.5 ounces (70 grams) of stars that were approximately 0.032inch (7.6 mm) in diameter, and approximately 0.64 ounces (18 grams) was rice hull break powder. http://www.jpyro.com...Kos-792-795.pdf

Approximately 2.68 inches (68 mm) in diameter and the shells had an average total mass of approximately 4.8 ounces (130 grams).

W = mg = 0.13kg x 9.8 m/s2 = 1.274N

Placing the mortar horizontally, and putting a loose fitting shell only (without lift charge) either a spherical shell, or cylindrical shell, and measuring the force to slide it down the mortar, will give us the total frictional force.

Similarly, if the shell has a snug fit, measuring the force to slide it down the mortar, will give us the total frictional force.

[digger]

You did not answer my questions.

 

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?

 

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. 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.

 

What is the relevance of the paper on shell failures (prevention of muzzle break)?

 

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.

[Crazy Cat]

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."

(1)

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

(2)

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)

Attached Thumbnails

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[Crazy Cat]

It's become clear to me Kenneth, that you, and Digger, are looking to smite and ridicule, before giving me the fair opportunity to post my complete work.

[digger]

Actually, I did, but you keep going tangent to the original commission of" for testing lift powders and tuning shell heights."
 

(1)
(2) 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.


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?

It has become equally clear to me, that you don't.


As I already have stated: External Ballistics: calculates, Reynolds number vs drag coefficient, Vertical Trajectory Calculations, & more. See attached picture.


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?


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)

 

OK, No you did not answer my question, I will try and put it more simply, forget about anything else. Why do you want to know what happens in the mortar if you have muzzle velocity? What is going on in the mortar that you want to know about and why? Please explain in simple terms for simple people like me to understand.

 

The reason the why the muzzle velocities/heights in the cricket ball tests varied so widely is simply down to the quality (burning speed) of the black powder. No more no less. Put simply, pressure rise, gas blow by, temperature from both burning and obviously from compression/expansion (not adiabatic due to blow by), incomplete use of the charge as still burning when ball has left the tube etc etc etc. It clearly has not a thing to do with the "external ballistics". (Lots of different black powders tested from many sources)

 

Whilst irrelevant, conventional Internal ballistic calculations generally rely on there being a seal between the object being fired and the barrel, even in the case of a rifled barrel. This is clearly not the case here. It just can't be treated to so simply to get a useful answer.

 

Hey I was only asking what the relevance of the paper was, so I guess it has a single example of a shell size and weight. Please bear in mind they do vary dramatically.

 

Oh and by the way what happens to the resultant force at the wall when your tube becomes vertical? oops it is 0, no friction. You will need to think that one out again, unless you are happy with that result for a ballistic friction.

 

By the way I am not looking to ridicule, you just have an extremely confrontational style and will not enter into discussion to work towards an answer that you need. I would have helped with the maths of what happens in the mortar just because I like deriving that sort of thing. I have £75K+ of CFD/FEM/DEM/symbolic mathematics software at my disposal that I use in my day job as chemical engineer (20 years experience) which can resolve it pretty quickly. But why bother you would not listen anyway, and just post a link to some other free bit of software that you have just searched for, which will be clearly way more suitable for the purpose.

 

Good luck with you endeavour let us know when you have finished and validated it.

Edited by digger, 24 November 2014 - 01:50 PM.

[Crazy Cat]

But why bother you would not listen anyway, and just post a link to some other free bit of software that you have just searched for, which will be clearly way more suitable for the purpose.

As you said, "What you need is evidence" I make reference to evidence, and you accuse me of the above.

OK, No you did not answer my question, I will try and put it more simply, forget about anything else. Why do you want to know what happens in the mortar if you have muzzle velocity? What is going on in the mortar that you want to know about and why? Please explain in simple terms for simple people like me to understand.

The reason the why the muzzle velocities/heights in the cricket ball tests varied so widely is simply down to the quality (burning speed) of the black powder. No more no less. Put simply, pressure rise, gas blow by, temperature from both burning and obviously from compression/expansion (not adiabatic due to blow by), incomplete use of the charge as still burning when ball has left the tube etc etc etc. It clearly has not a thing to do with the "external ballistics". (Lots of different black powders tested from many sources)

Whilst irrelevant, conventional Internal ballistic calculations generally rely on there being a seal between the object being fired and the barrel, even in the case of a rifled barrel. This is clearly not the case here. It just can't be treated to so simply to get a useful answer.

http://www.jpyro.com...ries/kosanke-2/

http://www.jpyro.com...Kos-138-149.pdf
http://www.jpyro.com...Kos-126-127.pdf
http://www.jpyro.com...Kos-167-172.pdf
http://www.jpyro.com...Kos-191-202.pdf
http://www.jpyro.com...Kos-203-215.pdf

By the way I am not looking to ridicule, you just have an extremely confrontational style and will not enter into discussion to work towards an answer that you need. I would have helped with the maths of what happens in the mortar just because I like deriving that sort of thing. I have £75K+ of CFD/FEM/DEM/symbolic mathematics software at my disposal that I use in my day job as chemical engineer (20 years experience) which can resolve it pretty quickly.

Who began with the insults and belittling: 10 June 2014. http://www.pyrosocie...ge-2#entry82257

"What a load of balderdash" "and is naive to think so"

Who has the extremely confrontational style??

I'm not going to post my credentials, but since I've already posted this on the Internet public forum, here's the link. http://www.bleepingc...s/#entry3541766

[digger]

Blah Blah

 

Can't you just type an answer to my question?, unfortunately I don't have the time to wade through endless links to try and decipher what you are eluding to, when two or three lines of your hand written text will do (by the way I have read a great deal of Kosanke's work in the past, which is very good).

 

So I will state it again, maybe you can manage to write something yourself this time.

 

Why do you want to know what happens in the mortar if you have muzzle velocity? What is going on in the mortar that you want to know about and why? Please explain in simple terms for simple people like me to understand.
 

Edited by digger, 26 November 2014 - 03:31 PM.

[digger]

brightened up my morning.

[Crazy Cat]

Blah Blah

Can't you just type an answer to my question?, unfortunately I don't have the time to wade through endless links to try and decipher what you are eluding to, when two or three lines of your hand written text will do (by the way I have read a great deal of Kosanke's work in the past, which is very good).

So I will state it again, maybe you can manage to write something yourself this time.

Why do you want to know what happens in the mortar if you have muzzle velocity? What is going on in the mortar that you want to know about and why? Please explain in simple terms for simple people like me to understand.

The predictable response I was expecting. I'm deliberately not answering your question, NOT to prevaricate, but to avoid feeding the troll. Your a clinical sociopath.

And so, you and your little minion Kenneth, and the sexual OCD you both share, and find amusing, can go happily into the sunset together.

Edited by Crazy Cat, 27 November 2014 - 05:05 AM.

[digger]

Aw thats the nicest thing anyone has ever said to me.

 

I'm off to go and kick the dog now.

 

By the way this thread does not seem to be developing anything useful at the moment so it is currently closed. If you want to get back on topic and get a sensible constructive discusion going on the topic at hand, let me know by PM and I will reopen the thread.