[roc-chat] Re: BP for Deployment of Main Question
- From: David.P.Smith@xxxxxxx
- To: roc-chat@xxxxxxxxxxxxx
- Date: Thu, 8 Dec 2011 12:08:27 -0800
Large areas can generate very large force
if you keep the pressure constant.
Think of it this way. For the
same length of tube, the diameter increases the area and volume by the
square of the increase in diameter.
Round off a 4 inch rocket tube to 12.5
square inches times 24 inches in length, the volume is 300 cubic inches.
Round off an 8 inch rocket tube to 50
square inches times 24 inches, the volume is 1200 cubic inches. Four
times the volume of the 4 inch.
So, if a 4 inch diameter body tube 2
feet long is pressurized to 20 psid with a given charge, then an 8 inch
body tube 2 feet long would be pressurized to 5 psid.
12.5 times 20 psid comes out to 250
pounds of ejection force. 50 times 5 comes out to 250 pounds of ejection
force.
So for a given ejection charge, in a
range of body tube diameters, you get the same ejection force.
On the other hand, increasing the charge
to four time the 4 inch charge, gives us 1000 pounds of ejection force
in the 8 inch rocket. that may be a bit much...
If you just stick to the formulae published
on some sites, you just keep increasing the charge size proportional to
the increase in volume, which may over stress your shock cord.
So how much force do you need to make
the rocket come apart?
Do you plan to ground test at all?
David P Smith
NAR 78668 L2
Amateur Extra, W6DPS
- The opinions expressed in this email are my own and do
not necessarily represent the positions, strategies or opinions of Southern
California Edison, its parent company Edison International, or any of their
affiliates.
|
From:
Richard Dierking <redierking@xxxxxxxxxxx>
To:
"roc-chat@xxxxxxxxxxxxx"
<roc-chat@xxxxxxxxxxxxx>
Date:
12/08/2011 11:36 AM
Subject:
[roc-chat] Re:
BP for Deployment of Main Question
Sent by:
roc-chat-bounce@xxxxxxxxxxxxx
Wow, 4 grams BP is about one half what we were coming
up using the web and a reference book I had.
Richard
Sent from my iPhone
On Dec 8, 2011, at 11:26 AM, "Chris J Kobel" <Chris.J.Kobel@xxxxxxxx>
wrote:
I would second David's recommendation
of 4 grams, which should give about 200 lbs of ejection force at 5000 AGL,
with the 3 2-56 shear pins requiring about 100 lbf to overcome.
Chris
From: David.P.Smith@xxxxxxx
To: roc-chat@xxxxxxxxxxxxx
Date: 12/08/2011
11:10 AM
Subject: [roc-chat]
Re: BP for Deployment of Main Question
Sent by: roc-chat-bounce@xxxxxxxxxxxxx
I use the "rule of thumb" on this site.
http://www.vernk.com/EjectionChargeSizing.htm
So, by that site you would need 4 grams of ffffg black powder.
Remember that you are looking at a lot of area to apply the ejection pressure
to. A 7.5 inch circle had a bit over 23.5 square inches of are. So
10 psid will give you 235 pounds of ejection force.
What is the shear force rating for your pins?
David P Smith
NAR 78668 L2
Amateur Extra, W6DPS
- The opinions expressed in this email are my own and do
not necessarily represent the positions, strategies or opinions of Southern
California Edison, its parent company Edison International, or any of their
affiliates.
|
From: Richard
Dierking <redierking@xxxxxxxxxxx>
To: <roc-chat@xxxxxxxxxxxxx>
Date: 12/08/2011
10:10 AM
Subject: [roc-chat]
BP for Deployment of Main Question
Sent by: roc-chat-bounce@xxxxxxxxxxxxx
Say your main deployment section is 7.5" diameter, 24" long and
using three (3) 2-56 nylon shear pins. How much BP?
Also, what's the advantage of using two or more deployment canisters instead
of just one? For multiple canisters, would you wire in series or
parallel? (One 9 volt battery and Perfectflite WD altimeter with
main deployment at 1100'.)
Richard Dierking
BTW: Kurt, we're not intending to turn the nose into a second stage :-)
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