A question for the engineers....

[Fat Elvis]

I have a question about egg drop stuff....

An egg typically weighs 55 grams.

It takes 25 Newtons to break an egg.

If you drop an egg from 25 ft, the egg will be travelling at ~40 ft/sec (12.2 m/sec) or roughly 27 mph when it hits the ground.

Assuming no use of a parachute or whatever to decrease the rate of fall, how much force needs to be "absorbed"/redirected or whatever to keep the egg from breaking?

How many Newtons (I'm not really sure what that is other than it is a unit of force where it is the amount needed to accelerate 1 kilogram of mass at the rate of 1 metre per second squared) would be exerted on the egg from a drop from that height?

I'm stumped.

[Fat Elvis]

OK, now I'm really confused. I've read online that it takes ~25 Newtons to break an egg, however, using the calculator in nstygma's link, you could drop a 55 gram egg ~5m before you'd reach that 25 Newtons threshold. I can drop an egg from a foot and in all likelihood it will break.

Can someone explain what I am missing here?

Thanks

[cdcox]

Originally Posted by nstygma

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

This calculator is very useful for answering the question. The key is the distance that the egg travels after impact. In case of a unprotected egg falling on a hard surface, the egg will travel a very short distance and the impact force will be very great and the egg will break.

But as you increase the distance travelled after impact, you are able to dissipate the energy and reduce the impact force.

According to the parameters you described in the OP, you would need to have enough cushioning that the egg could travel about 0.18 m after contact to reduce the impact force below 25 N.

[cdcox]

Originally Posted by Fat Elvis

OK, now I'm really confused. I've read online that it takes ~25 Newtons to break an egg, however, using the calculator in nstygma's link, you could drop a 55 gram egg ~5m before you'd reach that 25 Newtons threshold. I can drop an egg from a foot and in all likelihood it will break.

Can someone explain what I am missing here?

Thanks

I think you're looking at the Kinetic Energy K.E. You need to enter a stopping distance to get the impact force.

[saphojunkie]

[Consistent1]

Thought I was the only one who watched that Ryan Swope video. ha

[nstygma]

Originally Posted by Fat Elvis

OK, now I'm really confused. I've read online that it takes ~25 Newtons to break an egg, however, using the calculator in nstygma's link, you could drop a 55 gram egg ~5m before you'd reach that 25 Newtons threshold. I can drop an egg from a foot and in all likelihood it will break.

Can someone explain what I am missing here?

Thanks

there is so much variation in shell hardness, plus i think it can take a bigger impact at the tip of the egg compared to the sides.
so many varying answers online
http://www.ask.com/answers/239852281...o-break-an-egg

also, looks like the store-bought shells are weaker than fresh for some reason
http://www.backyardchickens.com/t/31...ard-egg-shells

[chop]

Just don't drop the eggs and you won't have to worry about them breaking on the floor.

[Exoter175]

The impact force is approximately 35.6818 Newtons, if the impact force required to break it is exactly 25 newtons, you'll have to avoid, mitigate, or redirect 10.6819 newtons to keep the shell from breaking.

[BlackHelicopters]

Wow. Weed!

[cdcox]

Originally Posted by Fat Elvis

OK, now I'm really confused. I've read online that it takes ~25 Newtons to break an egg, however, using the calculator in nstygma's link, you could drop a 55 gram egg ~5m before you'd reach that 25 Newtons threshold. I can drop an egg from a foot and in all likelihood it will break.

Can someone explain what I am missing here?

Thanks

OK I see your problem. You are using the default stopping distance of 0.1
m -- that is like 4". For an egg landing on a hard surface, the stopping distance is very short. If you use a stopping distance of 0.001 m, probably still too large, you will get a more realistic impact force, and a broken egg.

[cdcox]

Originally Posted by Exoter175

The impact force is approximately 35.6818 Newtons, if the impact force required to break it is exactly 25 newtons, you'll have to avoid, mitigate, or redirect 10.6819 newtons to keep the shell from breaking.

not correct.

You aren't considering the force generated by rapid deacceleration. Are you sure you are qualified to repair brakes?

[Eleazar]

Just put it in sawdust, I did that when we did this in 7th grade and it worked.

[Ace Gunner]

Originally Posted by salame

egg -- cellent

[Ace Gunner]

Originally Posted by Fat Elvis

OK, now I'm really confused. I've read online that it takes ~25 Newtons to break an egg, however, using the calculator in nstygma's link, you could drop a 55 gram egg ~5m before you'd reach that 25 Newtons threshold. I can drop an egg from a foot and in all likelihood it will break.

Can someone explain what I am missing here?

Thanks

times the square root of;



Then take the reciprocal

[Rausch]

Originally Posted by Cochise

Just put it in sawdust, I did that when we did this in 7th grade and it worked.

I remember doing this in 7th grade.

We were given specific rules on the weight of what we could use to stop the egg or protect it.

No rules were given about methods for catching it.

We put a big laundry basket under there with some really soft pillows and didn't even **** with the egg. We had outsmarted everyone.

My buddy dropped the egg and I think he was so sure we'd win he didn't really pay attention and the thing bounced off the side of the plastic laundry basket and smashed on the pavement.

We lose.


















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