Problem for physics, engineering, and other calculation hobbyists..
September 4, 2018 8:48 AM   Subscribe

I need to figure out equivalent drop force. There's a standard that says certain kinds of industrial glass must not shatter when a 2" diameter, 1.2 pound steel ball is dropped on it from a height of 20.25 inches. What I want to know is if I had a 20 pound object, at what (reduced?) height would I need to drop that object to generate the same impact force on the same surface? Thanks!
posted by soulbarn to Science & Nature (11 answers total) 2 users marked this as a favorite
I'm assuming that the change in contact area is not a factor in the specification. Otherwise finding a 2" diameter 20# object may prove difficult.
posted by Dr. Twist at 8:58 AM on September 4, 2018

I was just going to say the same thing -- that the question gets complicated if the shape (including curvature) of the object changes. A simple answer is, I think, going to depend on it being either a 2" diameter 20# ball, or at least a 20# object where the point that initially hits the glass has a 2" radius of curvature.
posted by LizardBreath at 9:03 AM on September 4, 2018

You'll need more information about the object, e.g. shape, hardness, orientation when it hits. The standard specifies a steel ball for a reason. A 20# lump of rubber won't have the same effect as a 20# steel cube.
posted by jon1270 at 9:03 AM on September 4, 2018 [2 favorites]

Otherwise finding a 2" diameter 20# object may prove difficult.

Not at all. An iron cylinder, rounded tip, 2" diameter and as tall as it would have to be to reach 20 lbs would satisfy that requirement.
posted by Stoneshop at 9:04 AM on September 4, 2018

OK, here's what I'm trying to find out, to make it clearer. The standard is for glass cooktops. So I want to know how that translates into the real world. So let's say that the 20 pound object is of the same material, and that it is 10" in diameter - so it's a spaghetti pot filled with water. It lands flat on the surface.

Does that help? Thanks!
posted by soulbarn at 9:09 AM on September 4, 2018

Agreed with all the above about contact area, coefficient of restitution, and lots of other things that will change; that’s why the test is standardized.

However, a physics 101 class would make an energy argument. The potential energy of an object is mass * height * gravity. If you want equivalent energy in your two scenarios, you need:
m1 * h1 * g = m2 * h2 * g
m1 * h1 = m2 * h2
m1 / m2 = h2 / h1

So, if you are increasing the mass of the ball by 16.67x (20lb/1.2lb), then the height needs to be decreased by the same factor:
20.25in/16.67 = 1.215 inches.
posted by RobotNinja at 9:13 AM on September 4, 2018

With the spaghetti pot landing flat I suspect the height would have to be much greater, because the force will be spread over ~78.5 square inches whereas the contact point of the ball has an area near zero. This isn't such a straightforward problem to solve.
posted by jon1270 at 9:16 AM on September 4, 2018

I don't think there's a clean answer for you. That is, it's going to depend on the rigidity of the pot, for one thing, which depends on the specifics of, e.g., how thick it is and so on. And it's also going to depend pretty importantly on whether it really does land flat-flat, or if it hits on a corner. Are the corners rounded or sharp? And so on.

Are you trying to figure out how careful you need to be with your cooktop? Because I'd take it as a rule of thumb that if that's the standard, it's probably good enough for normal use but don't push it -- literally dropping a spaghetti pot full of water onto a cooktop is a weird thing to do, so I wouldn't, but I also wouldn't sweat putting it down hard.
posted by LizardBreath at 9:17 AM on September 4, 2018

Your test ball will have a certain amount of energy, and a certain speed, the moment it hits the glass. This energy has to be transferred to the glass through an initially rather small contact area, which increases during the impact. The energy must be dissipated, and the speed reduced to zero in a very short distance (and time). This distance is what flexes the glass, and according to the specs it shouldn't break when deformed that amount. The steel ball also deforms a little during impact. Now your substitute object doesn't have the same physical properties regarding impact area and compressibility, so it'd be a serious slog through material property calculations to arrive at a meaningful result.
posted by Stoneshop at 9:29 AM on September 4, 2018 [2 favorites]

There are three quantities that would be of interest in an impact -- energy, momentum and force. You are not going to be able to make all three of these equal for a falling 2 pound ball and a falling 20 pound ball no matter how you set it up. For example energy is proportional to velocity squared and momentum is proportional to velocity. If you change the mass, it is impossible to produce both the equivalent energy and momentum at the same time. That is why testing standards prescribe a detailed precise repeatable test. It is the only way to replicate energy, momentum and force.
posted by JackFlash at 10:53 AM on September 4, 2018 [2 favorites]

At one startup I worked at we had a normal kitchen and a cook for ~20 people. Her largest pot filled with water just sitting on the stove while heating to a boil broke the glass top. Granted it was a large pot. But temperature at the time will be a factor too. What problem are you really trying to solve?
posted by jeffamaphone at 10:46 PM on September 4, 2018

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