Calculating the height of a spherical shell from flight time and shell design

OK, I checked it all out. You have macro's turned off.

 

I assume you are running a pre-2007 version of excel, go to the "Tools" Menu Scroll down to macro and then click on "security". Change the security level to medium. Then reopen the file and when prompted click enable macro's. This should solve your problem with the original file.

 

Hope this helps

 

Gareth

Can't do it then?

By the way I did say it was published and I did ask if you would like me to post it, but you decided it was not necessary.

The equations are simple, it just takes a bit of thought about the solution method.

Anyway for those who actually might be interested, see the attached shell height calculator. There are some variables that can be played with, for instance the drag table can be extended to improve number of data points.

I'm absolutely sure I can, and I did say "Maybe later on?". I don't need your calculator, but maybe the other members here did?

This ballisticscomputer.xls is excellent and can be modified too. ballisticscomputer.xls https://www.trijicon...icscomputer.xls
Scanned.https://www.virustot...sis/1414365024/

Since your running macros in your xls, I've scanned them through www.virustotal.com to check for malware. All ok, but not 100% accurate detection?
Results...
https://www.virustot...sis/1414364819/
https://www.virustot...sis/1414364891/
https://www.virustot...sis/1414364963/

 
I've look at your xls calculator today. Those of you using Open Office will fail to use the calculator, those of you using MS Office will function, but I found some issue with "shell height active x.xls" and "shell height form control.xls" - see snap shot.

Those of you who wish to begin to understand the formulas, look at page 8 & 9 of http://shellcalc.co....-Manual-610.pdf

Edited by Crazy Cat, 27 October 2014 - 06:54 AM.

As said this is an Excel file, no warranties for open office as I don't use it.

 

Nope no viruses/malware.

 

Oh and by the way ShellCalc whilst an excellent tool does not predict shell height and rise times based upon experimental dummy testing. It is intended for use with standard information about typical manufactured shells rather than useful data for experimenters to use to determine fusing/lift charge requirements etc. So keep searching the web, I am sure you will come across something that solves the question I posed eventually.

 

I look forward to seeing some original work from you soon.

nuff said as this isn't really the topic of this thread, and hence moved to its own thread.

 

 

 

Edited by digger, 27 October 2014 - 01:37 PM.

both above versions now seem to work using excel 2010

Thanks for letting me know Dave

 

Cheers

 

Gareth

Edited by digger, 27 October 2014 - 07:52 PM.

As said this is an Excel file, no warranties for open office as I don't use it.

Well maybe you don't, but many others do, as it's free and less vulnerable than MS Office. Open Office runs ballisticscomputer.xls and ShellCalc.xlsx without issue.

From Didier Stevens, a highly respected security researcher. #14. http://www.bleepingc...s/#entry3516000

The first file is a malicious .doc file.
The second file is a malicious .docx file with a VBA macro that executes a PowerShell script.

Since you are not using Microsoft Office on Windows, you run no risk, even if you opened the files with OpenOffice or LibreOffice on Linux.

OpenOffice or LibreOffice don't have the same vulnerabilities as Microsoft Office, and they don't support VBA.

As more vulnerabilities are discovered in Microsoft Office, October 21, 2014. https://technet.micr...ty/3010060.aspx | http://www.bleepingc...s/#entry3518065

Oh and by the way ShellCalc whilst an excellent tool does not predict shell height and rise times based upon experimental dummy testing. It is intended for use with standard information about typical manufactured shells rather than useful data for experimenters to use to determine fusing/lift charge requirements etc. So keep searching the web, I am sure you will come across something that solves the question I posed eventually.

You have taken it out-of-context. I was referring to the external ballistic formulas. Data on standard information about typical manufactured shells gives reference.

I look forward to seeing some original work from you soon.

I've started an XLS that predict shell height, rise times, based upon experimental dummy testing. Calculates mortar muzzle velocity, shell acceleration, and Newtons (N) of force acting on the shell or a give mortar length.

Now, where can I source "Internal ballistics" data for 'black gunpowder in X grams' = 'Y Newtons of force'

Now, where can I source "Internal ballistics" data for 'black gunpowder in X grams' = 'Y Newtons of force'

 

That is the problem; we are talking about experimental testing. That information has to be determined experimentally for a number of reasons. The acceleration of a shell over a mortar length will not only be determined by the burning rate of the BP, but also the fit of the shell in the tube. There is even a small dependence of burning rate linked to pressure (known as the pressure exponent), however this is small for black powder. The main problem here is that we are dealing with homemade BP hence the quality and speed will vary widely, this clearly has a huge influence on the force that it can impart on a shell. This is not even mentioning effect of grain size (Fg to 5Fg also FA and many other grades).

 

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).

 

In short that is why I picked flight time as the measured variable as it will give the most accurate results from the easiest to measure variable.

 

I guess if you want to solve the equations to give a useful (real world) result you should be using flight time as the measured variable also. You will need to use a numerical method to solve the equations as it is not possible to solve them analytically.  

At Digger,

Thanks for your input. I've sourced a voluminous amount of data and will post my finding as time permits. As with all experiments, and experimental data, there are always variables, and therefore, the data will have: averages, mean, mean deviation, etc.

In order to discover certain properties of a particular process or system, the investigators perform experiments or series of tests, in which the input variables of a process or system are changed purposefully such that from the observed output response the reasons for these changes could be identified. http://www.oru.se/Pa...course_plan.pdf

Statistics for Analysis of Experimental Data. https://www.princeto...chap_Peters.pdf

Some preliminary data. What's that 450Kg shell lift link again???

Attachment: 450KgShell.zip files contains...

shots.JPG
shot.txt
50cm radius at 450Kg Shell.pdf
100cm radius at 450Kg Shell.pdf
Convert.exe

Vertical Trajectory Calculation\quadrag.gif
Vertical Trajectory Calculation\quadrag2.gif
Vertical Trajectory Calculation\quadrag3.gif
Vertical Trajectory Calculation\quadrag4.gif
Vertical Trajectory Calculation\Vertical Trajectory Calculation.htm

OK, I have had a look at your calculation, I have not checked them thoroughly.

 

In the case of a large shell drag is less significant a force. However for smaller shell it does become quite significant.

 

I see you are assuming a uniform drag coefficient, as you have already stated this is not the case. You could resolve this by using either a drag coefficient table (in my opinion offers the best opportunity for accuracy, big table) or the Shimizu formula that is used by one of the companies in one of your links. In the former case you will have to use an iterative method to solve the equations. In the later case it may or may not be possible to solve the partial differential equation, but a simpler solution would also be to use an iterative method for solution.

 

How do you intend to calculate muzzle velocity to relate it to the real world?

 

keep it coming

Hello Digger,

In the case of a large shell drag is less significant a force. However for smaller shell it does become quite significant.

I see you are assuming a uniform drag coefficient, as you have already stated this is not the case. You could resolve this by using either a drag coefficient table (in my opinion offers the best opportunity for accuracy, big

table) or the Shimizu formula that is used by one of the companies in one of your links. In the former case you will have to use an iterative method to solve the equations. In the later case it may or may not be possible to solve the

partial differential equation, but a simpler solution would also be to use an iterative method for solution.

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.

How do you intend to calculate muzzle velocity to relate it to the real world?

I will be constructing an electronic muzzle velocity device prototype, and testing for accuracy as time permits; using a diode laser, 2 photoelectric cells + millisecond timing circuit.

However, there are also these options, considering the low muzzle velocities of 50 - 150 m/s.

Arrow Speed and Kinetic Energy. http://www.crossbown...etic-energy-r66

 
Radar Gun. http://en.wikipedia.org/wiki/Radar_gun | http://www.bushnell....ocity-speed-gun

Method for calibrating a muzzle velocity measuring device. http://www.google.co...tents/US4672316 | http://www.lens.org/...S_4672316_A.pdf

Device for determining the velocity of a bullet. http://www.google.co...4082670A1?cl=en

Low-Cost Optoelectronic Devices to Measure Velocity of Detonation. http://augean.elecen...5_proc_spie.pdf

Some input please?

Fg = mg = W (weight of an object is usually taken to be the force on the object due to gravity) in Newtons. http://en.wikipedia.org/wiki/Weight

W = mg.

A shell has a mass (m) of 0.3kg and gravity (g) is 9.8 m/s2. W = 0.3 x 9.8 = 2.94 N

 

(1) Mortars can be constructed from heavy cardboard, plastic (HDPE=High Density PolyEthylene), GRP (Glass Reinforced Plastic) or metal sunk into the ground or mounted in racks. The shell must fit snugly inside the mortar to allow proper thrust. Shells measure from two inches to three feet in diameter and can weigh up to 700 pounds. http://www.paramount...m/launching.asp

If shell, "must fit snugly inside the mortar to allow proper thrust" this is 2.94 N + frictional force between the shell and mortar surface + X newtons of force to accelerate the shell to a Y muzzle velocity.

So lifting force is, (Weight of shell 2.94 N) + (Friction N) + (Acceleration N) = Y muzzle velocity.


(2) The last thing to do now is to make sure the shell will drop freely into the mortar from which it is to be fired. If the shell is too snug, you can roll the leader side of the shell on a flat surface to reduce its thickness. http://www.skylighte...Ball-Shells.asp



If shell, "drop freely into the mortar" this is 2.94 N + X newtons of force to accelerate the shell to a Y muzzle velocity. Frictional force between the shell and mortar surface is negligible.

So lifting force is, (Weight of shell 2.94 N) + (Acceleration N) = Y muzzle velocity.

 

This also has effect on the ballistic chronograph accuracy, to measure Y muzzle velocity, of the shell exiting the mortar muzzle. No issue with "fit snugly inside the mortar", but the "dropped freely into the mortar" will cause the lifting charge expanding gasses + solids dust (smoke) to escape before the shell at the mortar muzzle. Usually, Infra Red (IR) LED's are used, but I was going to used a laser diode to compensate.

So my question is, "dropped freely into the mortar" or "fit snugly inside the mortar" ???

iv always found commercial shells the bigger the shell the looser they are in the mortar. 2" shell are very tight but 4" have noticeable gap. maybe smaller shells need more pressure to compensate for drag ect.

iv always found commercial shells the bigger the shell the looser they are in the mortar. 2" shell are very tight but 4" have noticeable gap. maybe smaller shells need more pressure to compensate for drag ect.

Thanks for the input.

Just a quick one

 

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.

 

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

 

The problem is the gap as you are aware.

 

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.