 # MiniSumo and Friction Forces

In this section we are considering Friction and how the minisumo interacts with the dohyo. Friction has a major effect on our ability to push as well as the resistance to being pushed. To simplify the maths standard units of measurement will be used for our calculations. First, considering our 'bot as a block on a level surface the reaction force (RN), from the surface, equals the gravitational pull to earth acting on the mass of the 'bot (mg). If we use the maximum weight of our 'bot then we have:-
 RN = mg = 0.5 kg x 9.81 m/sec2 = 4.9 kgm/sec2 This unit of kgm/sec2 is a bit of a mouth full, it's simplified in calculations to Newtons denoted by the letter N. RN = 4.9 N

To start moving this block across the surface we need to apply a force (P) which needs to overcome the frictional force that always opposes movement. The frictional force (F) is proportional to the reaction force (RN), if the mass doubles then the frictional force doubles.

F / RN = a constant (μ)

This constant is called the Co-efficient of Friction and varies for any surface material pairs, often called friction couples. The Co-efficient of Friction has two values, static and kinetic. Static friction, sometimes referred to as stiction, occurs before an object starts to move and is generally slightly higher than kinetic, or sliding, friction. Considering the example friction couples below we can start to consider our 'bot and interacting forces.

 Static (μs) Kinetic (μk) 0.25 - 0.5 0.2 - 0.3 0.9 0.7 0.2 - 0.6
Considering static friction on a wooden Dohyo with wooden and rubber based 'bots

F(wood) = μsRN = 0.5 x 4.9 N = 2.45 N

F(rubber) = μsRN = 0.9 x 4.9 N = 4.41 N

If we calculate again with kinetic values of μ it can been seen that once a 'bot starts to slide it takes less force to keep it moving. This model is a little too simple but applied to a more conventional 'bot shape it will help us see how to improve our pushing ability by sticking to the dohyo. Considering a simple 'bot shape, here there's a wheel to push on the rear and a skid/scoop on the front edge, the forces need a little thought. The contact surface area does not effect the overall frictional force but the balance of where the 'bots mass acts does. This 'bot shape can be simplified, for calculation purposes, to a simply supported beam as shown.

The total force acting up and down must be zero and the clockwise moments (clockwise turn effects of the forces) must equal the anticlockwise moments. Moments are measured as a Force Unit x Length Unit, i.e. Nm (Newton Metre).

If we spin about end A, then:-
Clockwise Moments = 0.03m x mg
Anti-Clockwise Moments = 0.08m x RB

As the beam is balanced:-
Clockwise Moments = Anti-Clockwise Moments
0.03m x mg = 0.08m x RB

Where for our 'bot mg = 4.9 N
RB = 0.03m x 4.9 N / 0.08m = 1.84 N

Now as RA + RB = mg (Vertical forces)
RA = mg - RB = 4.9N - 1.84N = 3.06N

Thinking of our minisumo in a stationary position we could consider frictional forces at the wheel and scoop to calculate how much force is required to push us out of the ring as well as the maximum push we could exert before spinning our wheels. Using the reaction forces, just calculated, and applying them to the earlier static friction calculations:-

FA = μsRA = 0.9 x 3.06N = 2.75N
FBs = μsRA = 0.4 x 1.84N = 0.74N
FBk = μkRA = 0.2 x 1.84N = 0.37N

From this we could say it takes as much as 3.49N to push us (FA + FBs), we can get as much as 2.75N push from our wheels only 2.38N is useful push (FA - FBk) as we still have to over come friction at the scoop. Note that if we where to try and push more than this 2.75N then the wheel would spin as we've over come the frictional force - this might happen if we're hard up against the opponent.

What hasn't been considered here is rolling resistance, this is a force that would resist the turn of a wheel. We have assumed surfaces are acting as if they are static while the wheel turns, if there is wheel spin then we'd be considering the kinetic friction.