What is the effect of articulation on handling? What is useful articulation?

Dear off-road and nature enthusiast friends, we are together with another sharing of our Off-road Basic Technical Information article series. The information I will provide; These are concentrated, technical-based, lived experiences, and some of the details that will find a name in the off-road literature for the first time. With the hope that it will benefit the readers.

Our topic today “What is the effect of articulation on handling? What is useful articulation?”:

Our topic is articulation, and a fine analysis that you won’t find anywhere else.

Let’s start with the basic definitions. You can find the definitions of articulation in the literature when you google it. Our friends who follow me know that I have made definitions and determinations independent of the literature in my articles. This time, I will describe it in the same way as I understand it from my own perspective.

Articulation capability is the ability of the wheel to move away from or approach the vehicle’s coupe. The greater this difference, the greater the articulation ability. We can call moving away positive articulation and approaching negative articulation . In general, positive articulation is more and negative articulation is less.

The answer to the question of why articulation is necessary will appear at the end of our article.

Let’s start from the very beginning, “the world was a cloud of gas and dust”.

The main force that makes vehicles move is the traction between the wheels and the ground. Traction is essentially a medical term. However, we will use it as we understand it. It is the traction force that occurs between the tire and the road and enables the vehicle to walk.

The more this force is, the better the tire can handle, and the better the tire can handle, the more successfully the vehicle can move forward. Whether this force is large or small; It depends on many parameters such as tire paste, pattern, width of the tread, tire air, the character of the ground on which one tries to move, etc.

Considering all these parameters, let’s assume that there is an “average coefficient of friction ” between the tire and the ground. The maximum value that the traction mentioned above can reach is the friction force between the tire and the ground. This frictional force is equal to the product of the average friction coefficient between the tire and the ground, which we mentioned above, and the normal force between the two surfaces, that is, the load transferred to the tread.

F_s = µ * N

There are two types of friction coefficient, static and kinematic . The static friction coefficient is the value before the slip starts, while the kinematic friction coefficient is the value after the slip starts between the surfaces. They can be easily determined by the inclined plane test . Since the kinematic friction coefficient has a lower value, the friction force, which is the upper limit of the traction force, will take a lower value after the skid starts.

In practice, it is sometimes possible to drive the car with increased skid. This is due to the fact that the interaction between the tire and the surface is not a uniform surface-to-surface contact, and the tire pattern and size are related to the average friction coefficient.

If we assume that the µ coefficient is constant for the moment we comment, within the parameters mentioned above, the main parameter that increases F_s is N . In other words, it is the share of vehicle weight per tire.

Keeping “µ” constant, if N increases, F_s increases, if N decreases, F_s decreases.

Now let’s examine why and how “N” increases or decreases. Our assessment is for spring suspension systems, but applies to all suspension systems within their parameters.

The task of the suspension is not only to provide driving comfort and / or safety. It is the system that transfers the load of the vehicle on the chassis to the axles and then to the wheels. That is, one task of the suspension system is to provide useful articulation .

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What is useful articulation ?
It is the state of articulation in which the weight of the vehicle on the chassis can be transferred to the axle as much as possible when the vehicle is articulated.

Let’s give a brief information about springs. There is a special coefficient (k) for each spring that determines the behavior of springs under load, that is, the character of the spring. This coefficient is the ratio of the force applied to the spring (F_y) to the amount of elongation or shortening (X) caused by the spring. (within the scope of Hook’s law)

That is, the force on the spring is equal to the product of the spring coefficient and the spring deformation.

F_y = k * X .

The force F_y on the vehicle spring is also the normal (perpendicular to the surface) force transmitted by the vehicle’s tire to the ground. In other words, the more load the springs in a vehicle suspension system can transfer to the tire, the higher the “N” within the framework of the basic rule 1 we mentioned above. The more “N” gets, the more F_s gets. The more F_s increases, the higher the upper limit of traction, the higher the upper limit of traction, the more successfully and in a large amount the vehicle can transfer the power from the engine to the road and move forward.

Now let’s examine the change in the force transmitted by the tire to the ground under off-road conditions. It will be more understandable to make this examination with numerical examples.

Suppose we have a vehicle with a total weight of 2200 kg. Let’s assume that the same amount of load goes to all four tires of this vehicle.

Let’s assume that the weight of the front and arc axles of this vehicle is 100 kg each.

That is, let’s assume that 2200/4 = 550 kg of load on each tire of the vehicle, 100/2=50 kg remaining on the axle (2200-2*100)/4=500 kg comes from the vehicle’s superstructure.

Let’s assume that the spring in the vehicle suspension system is 60 cm long when unloaded and shortens to 40 cm when the vehicle is loaded on it.

F_y = 500 kg

X=60-40=20cm

K = F_y / X = 500/20 = 25 kg/cm.

We can say that all movements that disrupt the distribution of the load transferred from the tires of the vehicle to the ground are falling diagonally.

We will evaluate the vehicle falling on the diagonal for 4 basic situations. All reviews are for the vehicle we have considered above. It has been assumed that there is no difference in the ground conditions of the individual tires and that all tires can transmit power within the same limit. If a situation contrary to this acceptance occurs, the new situation will display a different behavior within the framework of the same theories.

In this case, equal load will go to all tires.

N = 500+50 = 550 kg

F_s = µ * 500 kg.

As long as the traction can stay below F_s, the vehicle will be able to drive. If the power from the engine tries to transfer more force from the tires to the ground than F_s, all tires are expected to spin.

In this case, let’s assume that the arcs falling diagonally and extending are 50 cm, and the shortening arcs are 30 cm.

F_y_1 = K * (60-50) = 25 * 10 = 250 kg

F_y_2 = K * (60-30) = 25 * 30 = 750 kg.

In other words, only 250 kg of the vehicle weight will be transferred to the tire from the elongated springs, and 750 kg from the shortened springs.

N_1 = 250 + 100/2 = 300 kg

N_2 = 750 + 100/2 = 800 kg.

F_s_1 = µ * 300 kg

F_s _2 = µ * 800 kg.

It should be noted here that while the vehicle can transfer the same weight to all tires, the friction force, that is, the upper limit of traction, which is 500*µ kg, is 300*µ kg for two diagonal tires with extended springs. This means that these two tires will slip sooner. If you fall diagonally on a sloping ground, extra traction will be required in order to move the vehicle against gravity, thus reducing the upper limit of traction in this way may prevent the vehicle from moving. On the other hand, in the other two diagonal tires, the spring of which was shortened, the upper limit of traction was 800*µ kg. This means higher road holding capability for those tires. However, if the tires falling to the traction limit of 300*µ kg slip and slip, the vehicle cannot continue on its way if there is at least 1 differential lock.

In this case, let’s assume that the arcs falling diagonally and extending are 60 cm, and the shortening arcs are 20 cm.

F_y_1 = K * (60-60) = 25 * 0 = 0 kg

F_y_2 = K *(60-20) = 25 * 40 = 1000 kg.

In other words, only 0 kg of the vehicle weight will be transferred to the tire from the elongated springs, and 1000 kg from the shortened springs.

N_1 = 0 + 100/2 = 50 kg

N_2 = 1000 + 100/2 = 1050 kg.

F_s_1 = µ * 50 kg

F_s _2 = µ * 1050 kg.

It should be noted here that while the vehicle can transfer the same weight to all tires, the friction force, that is, the upper limit of traction, which is 500*µ kg, is 50*µ kg for two diagonal tires with extended springs. This means that these two tires will slip very, very soon. If you fall diagonally on a sloping ground, extra traction will be required to move the vehicle against gravity, so reducing the upper limit of traction in this way will prevent the vehicle from going. On the other hand, the upper limit of traction was 1050*µ kg in the other two diagonal tires with shortened springs. This means higher road holding capability for those tires. However, if the tires falling to the traction limit of 50*µ kg slip and slip, the vehicle cannot continue on its way if there is at least 1 differential lock.

Under these conditions, the axle self-weight will not contribute to the “N” weight calculated in case 3.

N_1 = 0 + 0 = 0 kg

N_2 = 1000 + 2*100/2 = 1100 kg.

F_s_1 = µ * 0 kg

F_s _2 = µ * 1100 kg.

It should be noted here that while the vehicle can transfer the same weight to all tires, the friction force of 500*µ kg, that is, the upper limit of traction, was 0 kg for two diagonal tires with extended springs. This means that these two tires are spinning empty. On the other hand, the upper limit of traction was 1100*µ kg in the other two diagonal tires with shortened springs. This means higher road holding capability for those tires. However, due to the empty wheels, the vehicle cannot continue if there is no at least 1 differential lock.

– Vehicles with high articulation ability fall on the diagonal more difficult, while vehicles with low articulation ability fall on the diagonal much more easily.

– All vehicles that are not exposed to articulation can drive the same way on straight roads, while vehicles with very short springs immediately on uneven surfaces It will fall into position 4. Vehicles with long springs, on the other hand, gradually take 2, then a very short moment. 3rd state and most recent They will fall into position 4.

– The situation is the same with scissor vehicles. The offroad earrings used to increase the articulation ability are placed on the 3rd floor of the vehicle. case it will take longer. As can be seen from the numerical example above, 3. The situation is not very positive in terms of traction. So we can define an equipment for the offroad tag that provides little useful articulation .

– Some vehicles also have anti-roll bar release systems. These are systems that provide benefits, provided that there is still a load on the springs, but 3 or 4. If the situation is reached, they will not show the expected positive effect.

– Suspension systems with very long soft spring coilovers that can be opened and closed in meters, make the vehicle much later. 3. and They are very efficient systems in the field as they reduce it to the 4th state.

– This article is the answer to why we should choose the longer shock absorber + spring option called the lift kit, rather than the chock, which is called the coupe lift, when making upgrade modifications to increase the terrain capabilities of the vehicles.

– We accepted that the vehicle that we examined the subject was a vehicle with a middle lock. Vehicle 2. or in case of slipping or Starting from the 4th states requires either at least 1 differential lock or an EDL-based traction support system . If neither of these systems is present, it will not be possible for the vehicle to continue on its way.

– Let me now connect the subject to another point that I always suggest.
When buying or modifying a vehicle, do not choose a vehicle without a center lock, at least the rear lock is also the front lock if possible, or the locks can be supplied easily and relatively inexpensively as additional accessories. Differential lock is one of the most important elements in off-road conditions.
Vehicle with differential lock + MT tire combination
The differential is generally more successful than the unlocked vehicle + XT tire combination.

Best Regards,

12/03/2020