How to calculate the force of an impact at a given speed (motorcycle protectors)?

Question

I am a motorcyclist, who knows next to nothing about physics :( , and was wondering if you could help me understand how to calculate the force of an impact, at a given speed:

if the rider hits a hard object, e.g. a wall, and comes to a sudden stop, and

if the rider hits the tarmac, then slides across the tarmac; in this case I’d like to understand the force of the initial impact with the tarmac.

The position of the rider, distance from the floor, lean angle, etc. will all be factors, but I guess some simplifying assumptions, just to get an idea of the order of magnitude we are talking about, would help.

I ask because European standard are very specific about the protective properties of motorcycle armour: a level 2 protector must transmit no more than 20 kN of energy when struck with a force of 50 Joules, (compared to 35 kN for Level 1). However, I have no clue whatsoever what these figures correspond to! I imagine the variables involved will be plenty, but I’d like to get a rough sense: a force of 50 Joules corresponds to… what? Kisisng the floor at 30 kmh? At 60?

All I have managed to calculate with my distant memories of high-school physics is that an impact at 50 kmh corresponds roughly to falling from a height of 10 metres.

I suppose hitting a wall at 50 kmh may easily kill you, but GP pilots kiss the floor at crazy speeds, slide and walk away unhurt, so I'm assuming hitting the floor and sliding involves much lower forces.

Answer 1

Ok first let's get the numbers clear.

$20kN$ is a force, not energy. $50J$ is energy, not force.

I think in your post you're talking about two scenarios. If you "fall off" your motorcycle and slide on the ground the danger is not so much the force slowing you down (which is basically only friction) but the friction damaging your skin, flesh, bones etc. If you hit a brick wall at a certain speed, you have kinetic energy $E_{kin}=\frac{1}{2}mv^2$, where $m$ is your mass, and $v$ your velocity. The wall will slow you down with very little space (and time) to do so, thus the force is extremly high. Protective equipment is on the one hand designed to endure the abrasive friction to protect your skin and on the other to reduce the force created by the impact, to protect your organs, mainly your spine and head.

On wikipedia it says that the damping properties are measured by dropping a 5kg cylinder on the protective equipment and measuring the transmitted force. You can calculate the energy of the cylinder as followed: $E_{pot}=m\cdot g \cdot h$ where $m$ is the mass of the cylinder, $g=9.81\frac{m}{s^2}$ the gravitational acceleration and $h$ the height, from which the cylinder is dropped.

Theoretically you can calculate the reactive force, if you know how much space you have to dissipate the kinetic energy of the cylinder: $E_{dis}=F\cdot d$, where $F$ is the force and $d$ the distance, but it is hard to predict $d$ as it depends on a variety of factors and material properties.

Answer 2

To answer this question we will need a few formulas:

• Kinetic energy: $E_k = \frac{1}{2} m v^2$; this is the energy tied to movement. $m$ is the mass of the moving object, $v$ its velocity.
• Potential energy: $E_p = m g h$; this is the energy an object can spend if descending from height $h$. $g = 9.81$ m/s$^2$.
• Newton's second law: $F = m\, a = m \frac{\Delta v}{\Delta t}$; this is the force a body of mass $m$ receives if it changes its velocity of a quantity $\Delta v$ in a time $\Delta t$. I assume the mass remains constant, otherwise the formula should change slightly. As a rule of thumb, a force of 10 N is roughly equivalent to having 1 kg laid on a body.
• Acceleration: $a = \frac{\Delta v}{\Delta t}$; you can obtain the number of "g"s by dividing that number by 9.81 m/s$^2$. Forces are measured in Newtons (N), energies in Joules (J). Other units will be expressed as SI units (kg for masses, m/s for velocities, m for lengths, s for times).

Let's talk about the first issue, the difference between hitting a wall and sliding on tarmac. For any kind of driver there are three main velocities to consider that can depend on the country you live but are all around the same values: city speed (50 km/h or 13.9 m/s), countryside speed (90 km/h or 25 m/s) and motorway speed (130 km/h or 36.1 m/s). For the rider I assume a mass $m$ = 80 kg. The mass of the motorcycle is irrelevant if the two separate during a fall.

Using the formula for kinetic energy, we can see easily that the energies for the three velocities are 7.7 kJ, 25 kJ and 52 kJ. As you can see, energy builds up quickly with the increase of velocity. This is the energy the rider alone has to dissipate if it stops, independently of the chosen method.

If you divide these three energies by $m\, g$ you can obtain the equivalent height of a vertical fall, without air friction. These heights are respectively around 10 m, 32 m and 66 m.

The real problem comes now. We talked about energy, but this isn't enough to understand what happens during a fall. For a better understanding we have to use forces. We know what $\Delta v$ is, it's the final speed (0 m/s) minus any of the speeds indicated above. Since we are not really interested in the direction of the force, we can neglect the fact that $\Delta v$ is negative and simply say that $\Delta v$ is the initial velocity. $\Delta t$ on the other hand is a little more tricky. It depends on the initial velocity both in the case of a collision with a wall and of the slide, but in different ways. To simplify a little bit the problem, i'll assume that it is instead constant. For the head-on collision the time interval is surely short, but not instantaneous. I'd estimate 0.05 s as an average. For the slide I watched a few videos on youtube and I think 5 s are adequate.

With these numbers we can calculate that for the collision the rider receives a force of 22 kN (28 g), 40 kN (51 g) and 58 kN (74 g) respectively, while for the slide the forces are 220 N (0.3 g), 400 N (0.5 g) and 580 N (0.7 g). I want to stress that these are rough estimates, but they should give you an idea of what these numbers mean. The first three are literally crushing forces, while the other three are probably only painful.

For the second part of your question, as Andrew already pointed out, some quantities don't seem to make sense. 50 J of energy seems a really small quantity if compared to a force of 20 kN. I can only suppose the energy involved is 50 kJ, this would be more plausible. In any case, the numbers given above, even if a crude estimate, should be enough to give you a basic understanding of what is going on.

Answer 3

Here is the directive is some more detail.

EN1621-1 Armor for All Body Parts (Except Back/Spine)

There are two European standards covering "motorcyclists' protective clothing against mechanical impact" - EN1621-1 and EN1621-2. EN1621-1 covers any body part protection except back/spine. EN1621-2 covers back/spine. Both standards assess the performance of protective devices by measuring the force transmitted through it when impacted by a falling mass.

EN1621-1 assesses armour designed to protect the shoulder, elbow and forearm, hip, tail-bone, knee and lower leg regions. The test apparatus consists of a mass of 5 kg with a 40 mm x 30 mm striking face, dropped onto the sample mounted on top of a 50 mm radius hemispherical dome. The anvil is further mounted onto a load cell, allowing a measurement to be made of the force transmitted through the protector. The kinetic energy of the falling mass at impact must not exceed 50 J.

A protector subjected to this test method is deemed to conform to this standard if the average transmitted force of nine tests is less than 35 kN, with no single test result exceeding 50 kN.

EN1621-2 assesses armour designed to protect the back/spine. It is a more stringent standard allowing no more than 18 kN of force to be transmitted to attain Level 1 protection (EN-1621-2 CE Level 1). Armour that allows less than 9 kN of force to be transmitted can attain a Level 2 protection (EN-1621-2 CE Level 2).

The directive is all about the transmitted force which in the case considered (body parts) must have as average transmitted force of nine tests of less than 35 kN, with no single test result exceeding 50 kN.

Before changes were made the Wikipedia quoted the kinetic energy as 50 kJ when it should have been 50 J and there was confusions between energy and force.
The value of kinetic energy sets an upper limit on the speed of the 5 kg falling mass and striking plate just before they hit the equipment under test. That speed is approximately 4.5 m/s or 10 mph.

This link provides much more detail about the test.

As with seatbelts what you are trying to reduce is the force which acts on your body.

When you are moving along you have a certain amount of momentum (mass times velocity). To reduce that momentum to zero a force must be applied for a certain time.
The relevant equation is "change in momentum = force $\times$ time".

If you have a mass of 50 kg and travelling at 20 m/s then to stop your momentum has to change by $50 \times 20 = 1000 \; \rm Ns$.

Hitting a solid object and stopping in $\frac {1}{100} \; \rm s$ would mean that a force of $1000 \times 100 = 100 \;\rm kN$ would act on you which could do some damage to you.
On the other hand if the time taken to slow down is increased to $\frac {1}{10} \; \rm s$ the force acting on you is now $10 \; \rm kN$ which would do less damage.
In part you protective motorbike gear increases time over which you slow down by being compressed on contact with the ground.
A recent way of increasing the slowing down time is to inflate the protective suits with a gas which then act like an airbag in a car.

Also to reduce the severity of any injury you should avoid rolling and try and slide so that you do not undergo multiple impacts with the ground.