The Tinetti Test

Text: Tim Hut

In the second part of the project of module 6 my group (consisting of me, Marion Fonseca Hoeltgebaum and Ties Martens) tackled a balancing problem regarding the Tinetti Test. The Tinetti Test [1] is used to assess the balance and stability of people in order to determine their fall risk. It consists of two parts, one related to the patient’s gait, and the other related to their balance while standing. We looked at the pushing section of the latter and in short it is: a patient is pushed lightly, if they can remain standing on their feet they get 2 point; if they need to take a step backwards they get 1 point and if they can do neither (and need to be stabilized by the physiotherapist) they get no point. This test has been determined to have quite a reasonable reliability [2], but The test can sometimes be impractical to apply, despite its low amount of necessary equipment and training, due to the potential discomfort of the patient, who might not enjoy being pushed, as well as the dangers of falling. Therefore, it would be beneficial to be able to simulate this test through modelling and predict the outcome using information about the patient which could be collected in some other way. Hence, our research question is: can we predict the outcome of the Tinetti Test based on the measured muscle strength of the patient?

Method First we start by creating a system of differential equations using the Euler-Lagrange equation. To represent a body, we used a three-link and a four-link, visible in figure 1a and 2a. The x and y coordinates of the endpoints of the kth segment of the n-link are rewritten in the form xk=x m +lksin(φk) and yk=ym+cos(φk), where m=k−1 if k≠n and m=2 if k=n. All variables related to the nth segment are often denoted with υ to signify the upper body. Next, the kinetic energy T, the potential energy U and the Lagrangian L are calculated.

Here mk is the mass at (xk, yk) and g is the gravitational acceleration. Next we use the Euler-Lagrange equation:

We rewrite it in the form:

Here the M represents the mass matrix depending on elements of φ and C is a column vector depending on elements of, among others, φ, where φ=[φ1 · · · φn]T . Extra torque Q is put into the system, therefore it changes into:

Let since matrix M is invertible, it follows that the system is:

If we would plot and animate the nlink for Q=0 it will collapse and fall through "the ground". Therefore the torque Q is made a feedback controller and is created such that the model convergences to a reference point. Since we have a nonlinear multiple input multiple output system, we used a method called feedback linearization. For this method it is required that the system is in companion form, that is for 1≤k≤n−1, and which is clearly the case for our system, if you let x=φ and u=Q. First the input state Q=h(φ,v,p) is found such that the system is transformed in an equivalent linear time invariant system: Q is chosen to be in the form b(φ)−1(p−f(φ,v)), thus Q=M(φ)+C(φ,v). Therefore, when this controller is applied, the obtained system is linear:

Now we can create another controller for p such that the poles of the characteristic polynomial are negative. That way, the system will become asymptotically stable. The controller is chosen to be in the form p=−α1φ−α2v for α1,α2≥ 0. And the characteristic equation becomes (s2+α2s+α1)n=0, which has negative poles. Lastly, to steer the system to the reference point the controller is adapted to include the values of the desired position, such that the desired position becomes the equilibrium. The new controller is:

Using this we write everything in one equation for Q,

Simulation To describe the entire movement, starting from an initial position we consider three phases after the push: returning to the start position, the swing phase for the step and the impact from the step with the ground. Each phase is adapted from the first system we introduced. Extra torque is added to represent the push and to allow proper knee bending since the human body has is restrictions how much it can bend and we made it proportional to the push. Let the total torque Q=Q1+Q2, then Q1 is the equation we last discussed and Q2 is in the form βH(−(t−t ))+γβ H(−(t−tγ)), where H(t) is the step function and β and γ are constants representing the energy by the push and the extra torque, which are active on the intervals [0, tβ] and [0, t γ] of time t. The person starts in an initial position standing straight for t<0 and is stabilized. At t=0 the push becomes active. We consider three possible events following from the phases coinciding with the scoring

of the Tinetti Test. The first is, a person is pushed and is able to return to the initial position modeled as a three-link. While the push is active the controller can’t stabilize to the equilibrium and upper body and the upper leg start rotating counterclockwise. At t=tβ the push becomes inactive and the person returns to the initial position. This is the first phase and succeeds if the torque from equation 2 doesn’t exceed the maximum torque. Then the person gains two points. If the first phase doesn’t succeed, the person needs to take a step backwards. Starting in the first phase it goes to the swing phase for the step as soon as the torque for the first time exceeds the maximum torque. The model changes to a four-link and the controller moves to a new equilibrium where it has one leg behind. Once again equation 2 is checked if the torques don’t go over the maximum torques. If the second phase succeeds it goes to take the impact with the ground. The last phase starts when the second leg for the first time hits the ground. The knee of the person is considered a spring, it stores energy as the knee is bent making the leg shorter, and it uses this energy to make the knee straight again. We use Hooke’s law, the spring force which acts on the movable side of the spring pushes the hip forward to a stable position: F=k(x−x0), to calculate an extra torque that acts on φ2. Here k is the spring constant and (x−x0) is the deviation from the length x0 of the spring and x is the length of the spring in meters.

References [1] Seifollah Jahantabi-Nejad and Qspring=Fspring×l2=kl2(x−l3). For the Akram Azad. Predictive accuracy of last time the torques are checked, performance oriented mobility asif the last phase succeeds the per- sessment for falls in older adults: son gets two points. If the second A systematic review. Medical Jouror the last phase fails, the person nal of the Islamic Republic of Iran, is unable to take a step and needs 2019. 10.34171/mjiri.33.38. to be caught by the physiotherapist. [2] M J Faber, R J Bosscher, and P Now comes the bad news, our mo- C W van Wieringen. Clinimetric prodel doesn’t predict the actual out- perties of the performance-oriencome of the Tinetti test. The main ted mobility assessment. Physical problem is that during our trials the Therapy and Rehabilitation Journal, torques used in the second and last 2006. https://doi.org/10.1093/ phase would always be higher then ptj/86.7.944. the torques in the first phase. This resulted that a person would always ether get 2 points or 0 points. It is unclear if that is due to overly high torque caused by our methods during swing and impact phase or due to overly low torque during stabilization, or both. But comparing with literature suggests the former.