Lift charge for shells

[d4j0n]

6g mini shell needs 4.5g lift to get to a decent 2 second apogee. Is this normal or is my lift really bad? Is there certain meshes that are best for certain sized shells?

[justanotherpyro]

Bp is weird like that. .1g won't lift a 1g shell in the same manor that 10g would lift a 100g shell. The difference in 2g of my good BP compared to 5g of the same BP was several hundred feet with the same projectile. 2g popped it out, and 5g launched it.

[d4j0n]

They say that there should be a few mm gap between the shell and the mortar...I assume this applies to larger mortars? A few mm's seems like alot for a 1.5" mortar...and I find that shells go higher in mini 3/4" and 1" mortars if the shells are snug from a wrap of tissue paper (I attatch lift under the shells with a tissue paper bag that is big enough to cover the shell as well).

[ApocalypsePlease]

Bp is weird like that. .1g won't lift a 1g shell in the same manor that 10g would lift a 100g shell. The difference in 2g of my good BP compared to 5g of the same BP was several hundred feet with the same projectile. 2g popped it out, and 5g launched it.

In a way, though, that makes sense to me. Here's my logic: For a larger shell, the free space between the shell and mortar is a very small percentage of the area of the shell. So, say you have a 2 mm gap on each side of a 10 cm shell. The ratio of free space to shell space is something like 2:25. (cylindrical shell, flat bottom)

However, seeing as how you still want the same amount of "free space" no matter what the shell size, if you had 2mm of space on each side of a 2.5 cm shell, the ratio is about 1:3.

 

But I don't know how this would play out in real life, like if the shell was snug in the mortar. It shouldn't matter then.

[Dr Boom]

The gap on the sides of a shell is somewhat critical- so in thinking that larger or heavier shells need more space, this is correct. You can make smaller shells "snug" such as in cake repeaters but when the aerials get bigger the pressures go way up. Try putting a 6 inch shell in a mortar that is 6 inches I.D. and you'll liking blow out the fuse, "bell" the round or have some other type of misfire. There has to be some gaseous escape before the shell moves or else it won't light.

 

In a gun we can see the effect of what we don't want in pyrotechnics- such high muzzle, breach, and cartridge pressures that everything must be thick heavy steel. In fireworks we can use a spiral wrap paper tube or (in heavy shell cases) plastic and HDPE.

 

If you have to use something so heavy in fireworks I should think you would be more interested in ballistics rather than amusing colored tapestries in the sky.

[weknowpyro]

This is a very useful document that is sure to help everyone. This was taken from the book 'the chemistry of fireworks' By M S Russell, So i will say thank you to him.

http://rapidshare.de/files/28534766/Aerial...l_tool.doc.html

[Pyroman2]

This is movie my Lift powder (BP grains, charcoal willow), amount 5 grams on the print paper.

Size 16-35mesh.

Test na kartce

 

( I use 6g my lift for 2" can shell who weight 80 - 100g)

[KevFla2001]

While testing out a new batch of bp in my 1" mortar today, I had an interesting thought. Does anyone know of a formula to convert seconds of airtime (using a dummy 1"od x 2-1/8, 23 gram shell) to altitude in feet? I can get 5.1 seconds from liftoff to ground strike using 7 grams homemade bp with that shell.

I'm guessing different size/weight shells would factor in to the formula but if anyone knows if one exists please tell!

Also, is the time factor equal in both directions? In other words, in my case, 2.5 seconds up, hang for .1 seconds or so in mid-air, then 2.5 seconds down? I would use this information to fine tune my fuse timing vs. lift charge.

[asilentbob]

You can find height with time and initial velocity... for one.

Meaning you would need a chronograph and a stopwatch.

 

Originally posted in "The Pyrotechnic Workshop Refrence"

The Pyrotechnic Workshop Refrence Thread

 

s(t) = -16(t)^2 + Vo(t) + Ho

t = Time in seconds

Vo = Initial velocity

Ho = Initial height

-or-

Y = -16x^2 + Vo(x) + Ho

Plug this into a graphing calculator and plug in your numbers to approximate things. Velocity is in FPS, initial height in feet, and the maximum y value will be the theoretical maximum height in feet. The graph will be a parabola. You can use these graphs to find alot of different stats.

 

ex. Y = -16x^2 + 200x + 0

A projectile fired at 200FPS (muzzle velocity) at an initial height of 0 (ground level). Once graphed we can see that the highest point on the graph (has max y value) is ~(6.25, 625), so at 6.25 seconds into the flight (exactly halfway) the projectile reached its theoretical max height of 625ft. We can also use this to estimate time fuse lengths for breaks, do you want a perfect break or do you want your shell to break throwing your stars up or down slightly.

SOURCE:

My College algebra course 2smile.gif. You can also find it on rec.pyrotechnics and many websites all over im sure.

[KevFla2001]

Thanks for the reply, and Merry Christmas!

 

I don't have a chronograph to determine initial velocity, and I could be wrong but I believe that initial velocity factors accurately only in a vacuum.

 

I know how to find height via visual angles and degrees, but I thought there might be a simpler way to determine height via hangtime that was specifically related to motars used in pyrotechnics.

 

Again, the question regards TIME. TIME to apogee vs. TIME to decent, in order to determine height and best fuse timing. If anyone can help I'd appreciate it.

Thanks to all

[styropyro]

A rough formula I use for determining height of shells is the simple "gravity makes everthing accelaerate downwards at 9.8m/s/s." I take the flight time from launch to back to the ground and divide it by two. So if a ball was in the air for 10s second I divide that by two to show how many seconds it took to reach its max height. Between the last second in the air it traveled between 4 and 5 seconds it moved about 5 meters because it was moving ~10 m/s at 4 seconds and 0 m/s at 5 seconds. Then I go down to the 3 to 4 second period in the air where it was moving ~20 m/s at 3 seconds and ~10 m/s at 4 seconds so it traveled 15 meters during that second period. I keep on going down until I get to the launch second then I add up all of the distances traveled in the seconds. So If it was a 5 second period to its max height it would have traveled 125 meters because 5+15+25+35+45=125. The only problem with this forumla is that There is no factor for wind resistance and if you blast something high enough there is nothing to factor in terminal velocity if it reaches it.

 

I hope this isn't too confusing

[Mumbles]

You're wrong about initial velocities only mattering in a vacuum, but it also doesn't matter here. Well, it matters, but it can be worked around to give a reasonably accurate. There have to be some unrealistic assumptions, but still it will be relativly accurate.

 

First we have to assume it went straight up and straight down. Also that said tests were done in a complete vacuum, or at least no air resistance or wind. Time up will be approximatly equal to time down. Time down has an initial velocity of 0. So.........

 

H= 1/2At^2

 

If you want meters use 9.81 for A, or if you want feet use 32 for A. t is of course half of your total time. So you get 67 feet.

[KevFla2001]

That's beautiful! Simple and easy to do!

 

It's been a long time since I've been in school so please tell me if my calculations are correct. Here go's.

 

5.1 seconds hangtime divided in half = 2.55

 

.5x32x2.55^2=104.04 ft altitude.

 

If that is correct and I'm doing the math right I've learned a very simple way to calculate altitude and fuse timing probably as accurately as possible. And it looks like my lift charge for the 1" shell is pretty close to optimum, would you agree? Or can you suggest something different? Thanks to all and Merry Christmas!!

[Mumbles]

Yeah, thats right, I messed up my calculation a tad. I used 2.05 instead of 2.55. Thats what I get for doing mental math right after waking up.

 

As for the timing thing, I still feel that a dummy test is the most accurate way. To get the height from this, you need to know the timing to the top of the apex in the first place.

[KevFla2001]

In order to calculate height using timing coming FROM the mortar, do i need to know muzzle velocity? Or is there a way to calculate going UP without that?

 

I appreciate your help... as always you are very well informed and helpful and I thank you for taking the time and effort to answer my questions.

[Mumbles]

To calculate height from the mortar to apex, yes you'd need to know the muzzle velocity, or at least figure it out. If you knew the force the lift placed on the shell you could calculate muzzle velocity. You can roughly calculate muzzle velocity from the hang time.

 

v= sqrt(2Ad) where A is acceleration due to gravity, and d is max height.

 

v= at where A is acceleration due to gravity, and t is time from apex to ground

[TheSidewinder]

Errr.... I thought that simply timing the shot from muzzle back to ground, dividing the time by 2, then using the resulting time to calculate a simple "freefall" of weight, ends up very close to correct.

 

e.g.

8 seconds round trip, therefore 4 seconds of "freefall".

4 seconds of ANYTHING (other than a single feather, e.g.) freefalling covers a known distance, 32ft/s/s^2, correct?

 

M

[KevFla2001]

Merry Christmas to All!!

 

I stumbled upon this link to the physics department at Gerogia State University last night and it completely blows me away, no pun intended!!

 

The program is called "Hyperphysics", and it not only covers trajectories as in the link below, but a click on the INDEX tab will bring up the entire program which includes dozens of other physics, chemistry, space, molecular, ect. programs, all with formulas and input boxes to automatically calculate your input data.

 

This is the trajectories page, and it thoroughly covers all types of trajectories just by inputing information into the boxes and clicking outside the box to compute.

 

Mumbles, I especially like the "Vertical Assent" section, as it will calculate the initial or "muzzle" velocity when height and time are imput, which we derive by the "freefall" section by inputing time to impact in decent, or by using the formula you provided yesterday, H= 1/2At^2.

 

Also please feel free to move this into a more appropriate section if it will help the members find it easier.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

[iris]

Hi,

I've just started making 3" shells utilizing the 2 5/8" diameter black plastic hemis. The finished dia. is also 2 5/8" dia. My finished shell weighs 92 grams excluding the lift charge. The mortar is 2 7/8" I.D. According to the rule of thumb (10% of the shells weight) the lift charge should be 150 grs. of black powder. This seems a little light. I use FFg. I see elsewhere a weight of 3/4 to 1 ounce (1 ounce = 430 grs.) is recommended. This seems too high. Can you give me some guidance on the correct weight of FFg black to use as a lift charge for my shell parameters?

 

Thank You

[busspuppy]

This thread is 7 years old. Ok just reread your post and you said 150 grains not grams. You confused me, try to stay in the same system if you can. Anways correct it would be around 6 grams. However remember that is for 2FA not 2FG . Also it dempends on how hot your bp is as well. I don't know for sure how much you will need, you might want to make a dummy shell to test. All bp is diffrent so it is hard to tell how much you need. Edited April 24, 2013 by busspuppy

[psyco_1322]

Do try to keep your weights listed as grams. In pyro, nobody uses grains, it's just too small of a unit. Some people don't know the difference and will assume you mean grams too.

 

10% of the shell weight is a good place to start. FFg, the grade for BP rifles is significantly finer than FFa, and therefore it does burn quite a bit faster. You would need more if you were using FFa, around about an ounce is common for 3" shells.

[pyrohacker]

I agree with frank you need to test a dummy shell.

 

But here are my specs:

 

Type of shell Sphere/Cylinder - ball

Shell size - 6"

Shell weight - 1600 grams

Wrapped or Unwrapped - What does this mean?

Tight or Loose fit in mortar - Neither, there is about 3/32" clearance

Lift charge Home made or Commercial - homemade

Amount of charge in grams - 106 grams

ADDED IN- grain size of lift - 6-12 mesh

Height of shell in air - 600-700 feet

 

Type of shell Sphere/Cylinder - Ball

Shell size - 4"

Shell weight - 360 grams

Wrapped or Unwrapped - ?

Tight or Loose fit in mortar - same as above

Lift charge Home made or Commercial - homemade

Amount of charge in grams - 24 grams

ADDED in - grain size of lift - 12-16 mesh

Height of shell in air - 400 feet

 

As you can see I use 1/15th the weight of my shell in powder. However, on a few recent shells that are unfired as of yet I have used 1/16th.

 

A good starting point is to use 1/16th for commercial, and about 1/14th for homemade if you believe your lift is good, 1/12 if its not so hot, and about 1/10 if it sucks. Make yourself a dummy shell, time it and figure the height, then adjust your lift accordingly.

 

Also grain size is important, thats why I added it in. I use 6-12 mesh for any shells 5" and bigger, OR a 4" shell thats quite large. and 12-16 mesh for 4" and smaller shells.

Referring to your post... I'm using #16 mesh size lift granules for my 2" shell and don't lift too much. Then I use #16 mesh for 1/2" little shell and it lifts high. Should I use a #10 or #12 to lift my 2" shells then? Any help would be greatly appreciated.

[MitchD55]

Ok I am new to pyrotechnics and I am starting to make some small can shells.I am currently using 2FA black powder for my lift but I can't seem to get it right.I was wondering if someone could tell me a good ratio for lifting small shells. By small shells I mean 1 to 2 inches

[lloyd]

Mitch,

First, 2FA is WAY too coarse for small shells. Most shells under 3" are lifted with 4FA or smaller (down to 7FA).

 

Second, the issues of mortar clearance and mortar length are issues. The less the clearance and the longer the gun, the higher the shell will lift with a given-amount of lift powder.

 

For a 2" shell that is a 'close fit' (say no more than 0.030" difference between tube i.d. and shell o.d. [not CLEARANCE, diameters]), about 6-8 grams of 7FA should be plenty. We lift 60mm shells from 14" guns with that amount.

 

(How's that for 'mixing units'? I can't help it. Our shells are metric, but the tubes are cut to Imperial!)

 

 

Lloyd

Edited May 10, 2017 by lloyd

[MitchD55]

Ok I'll try that thanks for the help