How to calculate the degrees of freedom of a simple multi-body system?

[davidoux2020]

Homework Statement

How to calculate the degrees of freedom of a tricycle when it rolls without slipping on the ground (a flat, horizontal rigid surface).

The tricycle is composed of a chassis, fork, and 3 independent wheels - 5 rigid bodies with perfect and play-free joints,

Relevant Equations

The general formula to compute DOFs in mechanical system is
DOF=3⋅(𝑁−1)−2⋅𝐿
with
𝑁 is the number of bodies
L number of joints

I intuitively understand that it has 2 degrees of freedom (rolling without slipping - RWS), but I struggle to formalize this according to the rules of the art:

I obtain:

10 - 3 (ground) - (rolling without slipping constraint) = 2

how to precisely calculate the RWS constraint? what formula and how to obtain 5 ?

 

[haruspex]

I am not sure exactly what is to be counted.
I would need to see the derivation of the formula. Is there a link you can post?

 

[kuruman]

My question is what exactly the general formula

DOF=3⋅(𝑁−1)−2⋅𝐿
with
𝑁 is the number of bodies
L number of joints

is counting and under what constraints. I can imagine a tricycle with its rear wheels welded on a common axle and the fork welded onto the chassis with the front wheel straight. I assume that I have 3 bodies: rear wheels that turn as one, chassis and front wheel. Applying the formula, I get

DOF = 3*(3-1) - 2L = 6 - 2L = . . . ?

Intuitively, I know that the tricycle has one degree of freedom: on a flat horizontal surface it can only roll along a straight line. Working backwards from this, I find $L=\frac{5}{2}$, an unacceptable half-integer answer. The simplest way to get DOF = 1 is with N = 2 and L = 1.

With the benefit of hindsight, I conclude that the rear wheels and front wheel should probably have counted as one "body" because they are constrained to move as one. However, "probably" is guesswork. I will not go into more guesswork and consider how to apply the formula when all welding is removed and the tricycle is returned to its pristine state.

I echo @haruspex's request to see how the formula is derived and how one counts the elements that go into it.

 

[haruspex]

I conclude that the rear wheels and front wheel should probably have counted as one "body" because they are constrained to move as one.

Yes, that works. It leads to the conclusion that, adding back a differential and steering, the answer to the original question is 3.

At https://support.ptc.com/help/creo/c...simulate/mech_des/calculating_dof_redund.html the formula is 6 (bodies) - constraints.
It does seem to me that 6 is right for a body in 3D space: 3 of position and 3 of orientation. That's one reason I was sceptical of the formula in post #1.

With the three body view your hobbled (no steering or differential) tricycle gives 6(2)-… what?
Since the front wheel has only one DoF relative to chassis, it must be considered to be under 5 constraints. That value is given at https://en.wikipedia.org/wiki/Degrees_of_freedom_(mechanics)#Mobility_formula.
Likewise the rear wheel pair. With one constraint from rolling contact, we have 6(2)-5(2)-1=1.

 

[davidoux2020]

Yes, that works. It leads to the conclusion that, adding back a differential and steering, the answer to the original question is 3.

At https://support.ptc.com/help/creo/c...simulate/mech_des/calculating_dof_redund.html the formula is 6 (bodies) - constraints.
It does seem to me that 6 is right for a body in 3D space: 3 of position and 3 of orientation. That's one reason I was sceptical of the formula in post #1.

With the three body view your hobbled (no steering or differential) tricycle gives 6(2)-… what?
Since the front wheel has only one DoF relative to chassis, it must be considered to be under 5 constraints. That value is given at https://en.wikipedia.org/wiki/Degrees_of_freedom_(mechanics)#Mobility_formula.
Likewise the rear wheel pair. With one constraint from rolling contact, we have 6(2)-5(2)-1=1.

Hi Haruspex, thanks for your message,
Actually I have the final solution of this problem but not the details , the solution "should be" 2 and
10 DOFs are used as initial value then reduced by constraints down to 2.
the final calculus (given) is 10 - 3[ground] - ( 6-1[Rolling Without Slipping Constraint] ) = 2
"it can only move forward and turn"

I initially struggled to come up with inital 10 DOFs then I realized the 3 wheels are independant (per problem statement), . I know its tricky and not realistic but its an, exam style problem with trickyness on purpose

the tricycle can be modelized by 2 bodies with 1 link,

To simplify lets start with the assumption the tricycle is in freefall, it has 10 DOF because :

6 DOF for the 2 linked bodies (3 Rot + 3Trans)
+3Rot (1 rot per wheel)
+1Rot the rotation axis of the wheel

Then it lands on a 2D rigid surface : -3 for the ground (minus 1 vertical axis, minus 2 rotation axises (yaw& pitch) which left 7 DOFs.

The remaining issue is I struggle to explain the (6-1) Rolling Without Slipping) even if i understand intuitively now how we come to 2 DOFs

 

[haruspex]

Hi Haruspex, thanks for your message,
Actually I have the final solution of this problem but not the details , the solution "should be" 2 and
10 DOFs are used as initial value then reduced by constraints down to 2.
the final calculus (given) is 10 - 3[ground] - ( 6-1[Rolling Without Slipping Constraint] ) = 2
"it can only move forward and turn"

I initially struggled to come up with inital 10 DOFs then I realized the 3 wheels are independant (per problem statement), . I know its tricky and not realistic but its an, exam style problem with trickyness on purpose

the tricycle can be modelized by 2 bodies with 1 link,

To simplify lets start with the assumption the tricycle is in freefall, it has 10 DOF because :

6 DOF for the 2 linked bodies (3 Rot + 3Trans)
+3Rot (1 rot per wheel)
+1Rot the rotation axis of the wheel

Then it lands on a 2D rigid surface : -3 for the ground (minus 1 vertical axis, minus 2 rotation axises (yaw& pitch) which left 7 DOFs.

The remaining issue is I struggle to explain the (6-1) Rolling Without Slipping) even if i understand intuitively now how we come to 2 DOFs

Both formulas are of the form A(N-1)-(A-1)(L).
The N-1 shows that we are to take one body as given, so all motion is relative to that.
The (A-1) implies that each linkage is assumed to remove all but one of the A degrees of freedom of a rigid component.
That works if the connectivity is a tree structure and the joints are like simple hinges. In that case, L=N-1 and both formulas give L as the answer.
It goes wrong when there are loops, and when a joint does not act like a simple hinge.

Consider four rods jointed with simple hinges in series. N=4, L=3, dof=3.
Now link the two end ones: dof=1 but the A=6 formula gives -2.
Or, instead, remove one link. dof=8, but the A=3 formula gives 5.

For the tricycle, both exceptions arise.

• Tyre contact with ground is like a universal joint, allowing rotation in two directions.
• There are two loops: chassis-fork-front wheel-ground-rear wheel-chassis (one such loop for each rear wheel).

Perhaps there is a classification of linkages such that a general formula can be made to work, but I have not got that far. For now, just go with the self evident fact that the vehicle can roll back and forth and the front wheel can steer: dof=2.

Btw, I stopped worrying about the differential. I realised that it serves to link the drive train to the rear axles. If we are ignoring the drive train as a body then we can discard the differential and take the rear axles as being independent.