Physics of the toilet displacement
This is a complex problem due to the elastic deformations of both the toilet and the body tissue of the person using the toilet. The amount of energy that is "imparted" into the system is equal to mgh (mass * height dropped * gravitational constant). For example, 100 kilos dropped 0.33 meters will impact with an energy of 323 Joules.
1 Joule = 1 Newton meter, or the amount of energy exerted when a force of one Newton is applied over a displacement of one meter. The critical part is the amount of displacement - if one can measure the amount of displacement (or deformation), one can extract the impact force. Notice that the smaller the displacement the greater the force.
Think of catching a falling egg. If you hold your hand still, all of the kinetic energy has to go into deformation of your hand and the egg = egg breaks into a nice big mess. If, instead, you "gently" catch the egg by moving your hand down, most of the energy is dissipated in the displacement = intact egg.
In the case of the toilet bolted to the wall or mounted to the floor, the object in question cannot be displaced. All of the energy, therefore, has to be absorbed through deformation of the toilet and the persons body tissue. It is very difficult to measure the deformation of either of these objects.
We therefore have to make some assumptions to get any sort of answer. I am in the process of analyzing the deformation of a porcelain object with respect to the known 1350 N (300 lb static load) failure limit.
In any event, I did some digging and found the results of a computer analysis of the impact force of a 20% supported (e.g. hands and feet) human body dropping on it's rear-end 2 inches. The object in question was the equivalent of a padded chair (allowing for a higher deformation that a toilet) and the answer was slightly over 2 g's.
So, using this value as a good (maybe great) ballpark estimate of the impact, we can get some quick answers.
1350 / (2 * 9.8) = approx 70 kilograms
So any unsupported weight exceeding about 150 lbs and dropped 2 inches onto a toilet is at risk of breaking it.
This is a linear answer, so basically multiply the unsupported weight by 2 for the equivalent static load when dropped from a height of 2 inches. |