# How Does Gravity Impact Engineering?

Leonard Kelley holds a bachelor's in physics with a minor in mathematics. He loves the academic world and strives to constantly explore it.

Gravity is quite important to engineering, and detailing all the reasons why would require books to delve into properly. So instead, look at these surprising examples that you might have not realized were true, yet are . . . .

## The Shape of a Bridge

Clearly, when you look at the cabling that runs between pylons of a bridge, we can see that they have a round shape to them. Though definitely not circular, are they parabolas? Amazingly, no.

In 1638, Galileo tested out what the possible shape could have been. He used a chain hung between two points for his work. He claimed that gravity was pulling the slack in the chain down to the Earth and that it would have a parabolic shape, or fit the line y2 = Ax. But in 1669, Joachim Jungius was able to prove through rigorous experimentation that this was not true. The chain did not fit this curve (Barrow 26).

In 1691 Gottfried Leibniz, Christiaan Huygens, David Gregory, Johann Bernoulli finally figure out what the shape is: a catenary. This name derives from the Latin word catena, or “chain.” The shape is also known as a chainette or a funicular curve. Ultimately, the shape was found to result not only from gravity but from the tension of the chain that the weight caused between the points it was attached to. In fact, they found that the weight from any point on the catenary to the bottom of it is proportional to the length from that point to the bottom. So the further down the curve you go, the greater the weight that is being supported (27).

Using calculus, the group assumed that the chain was of “uniform mass per unit length, is perfectly flexible, and has zero thickness.” Ultimately, the math spits out that the catenary follows the equation y = B*cosh(x/B) where B = (constant tension)/(weight per unit length) and cosh is called the hyperbolic cosine of the function. The function cosh(x) = ½*(ex + e-x) (275).

## Designing Roller Coasters

Though some can view these rides with great fear and trepidation, roller coasters have a lot of hard engineering behind them. They have to be designed to ensure maximum safety while allowing for a great time. But did you know that no roller coaster loops are a true circle? Turns out if it were that the g forces experience would have the potential to kill you. Instead, loops are circular and have a special shape. To find this shape, we need to look at the physics involved, and gravity plays a big role (Barrow 134).

Imagine a roller coaster hill that is about to end and drop you off into a circular loop. This hill is a height h tall, the car you are in has mass M and the loop before you has max radius r. Also note that you start higher than the loop, so h > r. From before, v2 = 2gh so v = (2gh)1/2. Now, for a person at the top of the hill all the PE is present and none of it has been converted to KE, so PEtop = mgh and KEtop= 0. Once at the bottom, that entire PE has been converted to KE, to PEbottom = 0 and KEbottom = ½*m*(vbottom)2. So PEtop = KEbottom. Now, if the loop has a radius of r, then if you are at the top of that loop then you are at a height of 2r. So KEtop loop = 0 and PEtop loop = mgh = mg(2r) = 2mgr. Once at the top of the loop, some of the energy is potential and some is kinetic. Therefore, the total energy once at the top of the loop is mgh + (1/2)mv2 = 2mgr + (1/2)m(vtop)2. Now, since energy can neither be created nor destroyed, the energy must be conserved, so the energy at the bottom of the hill must equal the energy at the top of the hill, or mgh = 2mgr + (1/2)m(vtop)2 so gh = 2gr + (1/2)(vtop)2 (Barrow 134, 140).

Now, for a person sitting in the car, they will feel several forces acting on them. The net force they feel as they ride the coaster is the force of gravity pulling you down and the force the coaster pushes up on you. So FNet = Fmotion (up) + Fweight (down) = Fm – Fw = Ma - Mg (or mass times acceleration of car minus mass times acceleration of gravity) = M((vtop)2)/r – Mg. To help make sure that the person will not fall out of the car, the only thing that would pull him out would be gravity. Thus the acceleration of the car must be greater than the gravitational acceleration or a > g which means ((vtop)2)/r > g so (vtop)2 > gr. Plugging this back into the equation gh = 2gr + (1/2)(vtop)2 means gh > 2gr + ½(gr) = 2.5 gr so h > 2.5r. So, if you want to reach the top of the loop courtesy of gravity alone, you much start from a height greater than 2.5 times the radius (Barrow 141).

But since v2 = 2gh, (vbottom)2 > 2g(2.5r) = 5gr. Also, at the bottom of the loop, the net force will be the downward motion and the gravity pulling you down, so FNet = -Ma-Mg = -(Ma+Mg) = -((M(vbottom)2/r + Mg). Plugging in for v bottom, ((M(vbottom)2)/r + Mg) > M(5gr)/r + Mg = 6Mg. So when you get to the bottom of the hill, you will experience 6 g’s of force! 2 is enough to knock out a kid and 4 will get an adult. So how can a roller coaster work? (141).

The key is in the equation for circular acceleration, or a c = v2 / r. This implies that as the radius increases, the acceleration decreases. But that circular acceleration is what holds us to our seat as we go across the loop. Without it, we would fall out. So the key then is to have a large radius on the bottom of the loop but a small radius on the top. To do this, it must be taller than it is wider. The resulting shape is what is known as a clothoid, or a loop where the curvature decreases as the distance along the curve increases (141-2)

## Works Cited

Barrow, John D. 100 Essential Things You Didn't Know You Didn't Know: Math Explains Your World. New York: W.W. Norton &, 2009. Print. 26-7, 134, 141-2, 275.

This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.