41 replies to this topic

[Atom Fireworks]

Depends on the size of the shell I guess. 100ft+ sounds good for a 3" shell.

Yea 3 inch, and also i have had 1 did not light upon launch it did not land until 12 seconds later, the shell weighed 200 grams is there a ways to work out how high it went?

Cheers Dumper

[portfire]

Just to throw some numbers in. The width of strip should be 1/16 the circ of the OD of the shell and the length 1/3.
Bit late but there you go, hope it helps

[CCH Concepts]

Yea 3 inch, and also i have had 1 did not light upon launch it did not land until 12 seconds later, the shell weighed 200 grams is there a ways to work out how high it went?

Cheers Dumper

I made it about

Assuming 2 seconds flight time so 10 second fall time

Terminal velocity seems to be 70 m/s/s using




Assuming spherical 75mm shell

This would be reach after 7.14 seconds

Now taking distance travelled as



And



Average velocity = 35 m/s

Distance = 250m

This now leaves the distance travelled at Vt

2.86 sec at 70 m/s

Another 200m

Total 450m

Now this seems very high, so I’m not sure if my maths is right hence me showing the workings, mechanics was never my strong subject.

Edited by CCH Concepts, 18 December 2009 - 01:07 AM.

[Atom Fireworks]

I made it about

Assuming 2 seconds flight time so 10 second fall time

Terminal velocity seems to be 70 m/s/s using




Assuming spherical 75mm shell

This would be reach after 7.14 seconds

Now taking distance travelled as



And



Average velocity = 35 m/s

Distance = 250m

This now leaves the distance travelled at Vt

2.86 sec at 70 m/s

Another 200m

Total 450m

Now this seems very high, so I’m not sure if my maths is right hence me showing the workings, mechanics was never my strong subject.

Well in all honesty i used 40 g bp lift so 20% of the shells weight just because i dont want a shell burstin to low or even on its way back down, its worked for the other 20+ that ive fired sucessfully. CCH ime not good with maths so this might seem a daft question at what hight do u believe it reached its apex? 200 or 450 meters,

Portfire cheers for heads up on strip dimensions al be sure to use them next time i past anything up hopefully in the crimbo holidays maybe a cheeky 4 or 5 inch willow as i now have the proper chinese time delay fuse ,

Thank you for this info guys its invaluable but could somone back up CCH.

Bumper

[digger]

Well in all honesty i used 40 g bp lift so 20% of the shells weight just because i dont want a shell burstin to low or even on its way back down, its worked for the other 20+ that ive fired sucessfully. CCH ime not good with maths so this might seem a daft question at what hight do u believe it reached its apex? 200 or 450 meters,

Portfire cheers for heads up on strip dimensions al be sure to use them next time i past anything up hopefully in the crimbo holidays maybe a cheeky 4 or 5 inch willow as i now have the proper chinese time delay fuse ,

Thank you for this info guys its invaluable but could somone back up CCH.

Bumper

OK I have not thought the math through myself yet. But my initial thoughts are that there is a flawed assumption in there. The rule of thumb is 1/3 of flight time on the up (maybe a little more) and 2/3 of flight time on the down so it should be 8 seconds fall and not 10 for starters.

I will check the math over the weekend.

P.S. 450m is an unreasonable answer.

Edited by digger, 18 December 2009 - 07:56 AM.

[CCH Concepts]

yer thats was one of the main things i wasn't sure over.

so that would be 7.14 to Vt to give 250m and 60 meters at Vt. so total of 310m.

although this still seems very high.

digger i will be interested to see if you can see where i went wrong or confirm my maths.

[Atom Fireworks]

310 meters is high isnt it. if it went that high i will be proud of my b.p for sure. Digger can you confirm cch's hight or have you drawn a different conclusion.

Dumper

[digger]

Right I have taken a quick look at your maths CCH.

There are a few problems. You can't work out the distance traveled using an average velocity because you are effectively ignoring drag, as this changes as a cube of the velocity.

The easiest way to model this is to do a force balance and then solve iteratively (because the drag effects the acceleration).

To solve it will be necessary to guess the muzzle velocity and then solve for height and time of flight. It is a simple matter to change the muzzle velocity with a circular iteration until the time of flight matches the actual. This will give you the height of the shell. One thing to note it is important to check the Reynolds number because if it exceeds 5 x 10^5 (about 150m/s) then you get boundary layer separation which dramatically reduces the drag coefficient. This is why golf balls are dimpled to bring on boundary layer separation at a lower velocity (increase the Reynolds number).

Anyway I will knock together a spreadsheet this evening that does the calculation and post it up here. I think you will find it gives much smaller answers than your model.

[Atom Fireworks]

Right I have taken a quick look at your maths CCH.

There are a few problems. You can't work out the distance traveled using an average velocity because you are effectively ignoring drag, as this changes as a cube of the velocity.

The easiest way to model this is to do a force balance and then solve iteratively (because the drag effects the acceleration).

To solve it will be necessary to guess the muzzle velocity and then solve for height and time of flight. It is a simple matter to change the muzzle velocity with a circular iteration until the time of flight matches the actual. This will give you the height of the shell. One thing to note it is important to check the Reynolds number because if it exceeds 5 x 10^5 (about 150m/s) then you get boundary layer separation which dramatically reduces the drag coefficient. This is why golf balls are dimpled to bring on boundary layer separation at a lower velocity (increase the Reynolds number).

Anyway I will knock together a spreadsheet this evening that does the calculation and post it up here. I think you will find it gives much smaller answers than your model.

Cheers digger cnt wait to see what you come up with.

Dumper

[CCH Concepts]

when i first started at uni i took mechatronics. this is why i changed to electronics, lol. there are so many variables, just when you think you have something right, someone mentions a rule, phenomena or effect you have either not considered of even never heard of. i didn't consider drag, although drag was used to calculate the Vt. as for the rest its very obvious digger is by far and away better at this than me. ill be very interested to see what he does.

i did think the average v was my down fall, i thought i was going to need an integral to get a more exact answer, guess i would have still been wrong.

are you going to use some sort of matrix with varying V. this Reynolds number number sounds interesting, never heard it refereed to before, or i did and it went over my head.

[digger]

Right then I have done a spreadsheet that does the calculations. It can be found here Click Me.

The sheet uses macros so these need to be enabled. Change the items in green to calculate your shell flight. Change the items in blue if you know what the code does.

The code is rough and ready so silly values in the green boxes won't resolve and it will lock in a terminal loop. I will sort this out if I get time at some point.

I will upload an explanation of the maths over the next week or so.

A few interesting points I noticed whilst doing the program are:
1.)The smaller shells will go into the low drag coefficient range with high Reynolds numbers.
2.)The smaller shells do not approach terminal velocity before they hit the ground!
3.)The larger the shell the up and down times start to approach equal time!

Anyway the answer to the problem is just below 170m assuming that the shell weight was 150 grams.

Well have a play and look at the code to see if you can find any errors or even optimise it (plenty of scope for optimisation).

[Atom Fireworks]

Right then I have done a spreadsheet that does the calculations. It can be found here Click Me.

The sheet uses macros so these need to be enabled. Change the items in green to calculate your shell flight. Change the items in blue if you know what the code does.

The code is rough and ready so silly values in the green boxes won't resolve and it will lock in a terminal loop. I will sort this out if I get time at some point.

I will upload an explanation of the maths over the next week or so.

A few interesting points I noticed whilst doing the program are:
1.)The smaller shells will go into the low drag coefficient range with high Reynolds numbers.
2.)The smaller shells do not approach terminal velocity before they hit the ground!
3.)The larger the shell the up and down times start to approach equal time!

Anyway the answer to the problem is just below 170m assuming that the shell weight was 150 grams.

Well have a play and look at the code to see if you can find any errors or even optimise it (plenty of scope for optimisation).

Cheers digger, the shell actually weighed 198 grams and then i used 40 g lift, the flight time was 12 seconds from launch to land. I have clicked on ur link and my computer wont let me open it up but am crap with computers anyways ( Bricky by trade). Also guys i thinking ov giving a 5 inch willow a go. does the same 15% lift apply or is trial and error.

P.s you help is much appreciated digger and cch thank you very much.

Dumper

[digger]

Save the file to your hard drive first and then open it, it should work fine then.

For those Variables you get
Apogee of 168.9m
Time to Apogee 5.08s
Muzzle Velocity 76.6 m/s (171.4 mph)
Impact Velocity 37.6 m/s (84.1mph)

Based on those numbers I would be tempted to back off your lift to about 25g as your shell would still be on the rise when it went off. I guess you should be aiming for 3 - 4 seconds max to apogee.

Edited by digger, 22 December 2009 - 10:55 AM.

[Mumbles]

I have no clue where you got the 15% lift charge as that is WAY overkill. If you have to use more than 10%, your lift really is just inferior. If you are confusing 15% with 1/16, you may need to take a closer look at the numbers.

[Atom Fireworks]

I have no clue where you got the 15% lift charge as that is WAY overkill. If you have to use more than 10%, your lift really is just inferior. If you are confusing 15% with 1/16, you may need to take a closer look at the numbers.

I have read somewhere that with home made bp you should be using around 15%. If using the real deal then 1/10 or 10% should be used. My lift is probably not the fastest but i like the idea of too much lift as opposed to too little.

So digger would you recomend trying 25g per 200g shell looking at those numbers?

Dumper