NAR Event in 2007!

[Bret PackardParticipant]

I’ll look forward to frying up the aftermath for a late prarie breakfast 😀

[denverdoc]

I’ll look forward to frying up the aftermath for a late prarie breakfast 😀

Hey you can have breakfast but, here’s the deal, if I do recover the egg intact, I get a photo op with egg on your face that I can submit to TRF 😛
JS

[Bret PackardParticipant]

Sounds good, it would be worth it to see an egg survive a W9 load. Should be fun to watch, good luck!

[Chris LaPanse]

I’ve got a system worked out, and I think it should survive (after some feasibility drop testing on the deck just now 😆 )
Heck – I may even spring for 54mm W9 😯

[Chris LaPanse]

Just did some more drop testing outside for my latest method of egg-protection – got up to at least a few hundred gees impact (15 foot drop onto concrete). Something completely unexpected happened – I cracked the tupperware, but the egg survived. I think I have my method of containment figured out 😀

Also, if it’s just for fun rather than as a contest, I’ll go for the J1999 instead – my egg can handle it 😯

EDIT: just cracked open the egg – stunningly, after an impact of at least 350 gees, possibly more, the yolk was 100% intact.

EDIT #2: After more calculations, it would appear that it actually withstood an impact acceleration of around a thousand gees, though admittedly only for about a millisecond

[Warren B. MusselmanModerator]

How did you figure the G load? Accelerometer and a computer data acquisition system?

Warren

[Chris LaPanse]

Nope. Simply knowing the speed at impact and estimating the distance that it stopped in based on the amount of deflection needed to cause a crack of the kind observed in the tupperware container. Turned out to be about a sixth of an inch to stop from 9.6 meters per second to zero. That’s about a thousand gees. Even if it were to take a half inch to stop (MUCH longer than it really could have), that’s 360 gees. Basically, it’s far more acceleration than the W9 would create, and 1000 gees seems the most accurate estimate I can make with the current information.

[Warren B. MusselmanModerator]

What’s the math involved in this? Show me the formula since I must admit to forgetting all that stuff I learned in college – I finished engineering school (at least the ME portion) before you were born.

Warren

[Chris LaPanse]

I used the time-independent acceleration equation, V(f)^2-V(i)^2 = 2ad

In this case, after a fall of around 15 or 16 feet (around 4.9 meters), the velocity works out to be right at 9.8 meters per second, with a fall time of 1 second. The impact distance is around 1/6 of an inch, which is 0.0042 meters. So, we get (0)^2 – (9.80)^2 = 2*(a)*(0.0042)
This simplifies out to -96.04 = .0084*a
Divide both sides by 0.0084 and you get an a of 11,433 m/s^2. Divide this by 9.8 and you get 1167 gees.

(Apparently, AP physics is good for something after all 😉 )

[Bruce R. Schaefer]

Checked out CRASH’s discussion on NCR’s NAR event. The E booster glider seems to be getting a good response. But the G Super-Roc seems to be getting a bad response because of safety. Does anyone agree or disagree with this? While you can’t use fiberglass tubes, you CAN glass paper tubes. Dale, if you read this, what are your thoughts? Joe, Warren, safety? Sending a spiraling, collapsing G into the flight line, even if pointed away from the crowd, is a possibility. We can certainly start whittling this down and dropping the larger events, if need be. Since not many have responded, there is still the option to make this a totally kid NAR event.

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