Torticollis is a condition where an infant's neck muscle is tight or shortened on one side, causing the head to tilt in one direction. Congenital muscular torticollis (CMT) is the most common form and is typically identified in the first few months of life. Physical therapy with gentle stretching, positioning strategies, and parent education is the primary treatment.

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Yes. Lead Pediatric Physical Therapy is an in-network provider with TriWest Health Care Alliance, which administers TRICARE West. Your provider can call 844-866-9378 to enter a referral. We are honored to serve military families in Fort Worth.

Do you offer in-home physical therapy visits?

Yes. In-home and daycare visits are available within a 15-minute drive of our clinic in the 76107 area of Fort Worth. Your child receives the same one-on-one care with Chrissy in the comfort of their own environment.

What is a PCS certification?

PCS stands for Pediatric Clinical Specialist, a board certification awarded by the American Board of Physical Therapy Specialties. Fewer than 3 percent of physical therapists in the United States hold this credential. It requires advanced clinical experience and a rigorous examination focused exclusively on pediatric physical therapy.

How much does a pediatric physical therapy evaluation cost?

The initial evaluation is $185 and lasts 45 to 60 minutes. Follow-up one-hour treatment sessions are $155 and half-hour sessions are $95. Packages are available at a reduced rate. We accept HSA and FSA cards, credit and debit cards, and Venmo.

What should I bring to my child's first physical therapy appointment?

Bring any referral paperwork from your pediatrician, a list of your concerns and questions, and your child in comfortable clothing. If your child is an infant, bring a favorite toy or comfort item. Chrissy will spend time understanding your child's history and your goals before beginning the evaluation.

Engineering & Code

Garage Door Simulator

by Tim Booher October 10, 2026

My son had a physics assignment to estimate the power required to raise a garage door. He needed to make a video for this. I wanted to build an interactive simulation to teach about a simplified linearized model versus what differential equations can actually do in the real world.

Now that I’m using AI to understand the world around me, my son’s physics project provided an opportunity to understand how garage doors work. They’re super simple, just a motor that pulls on a door. But there’s a couple interesting things going on. The first one is that it’s possible when you disconnect the motor to just lift the several hundred pounds of door up with a single arm because the torsion spring is pulling it up. And the door starts to move horizontally after the door goes around the first turn, which gets it out of the way but also makes it easier to move.

Illustration 1 – Drag the slider to lift the case off, then run the motor and watch the worm, gear, sprocket and chain open the door.

The motor itself in my garage runs on 120 volt AC and spins really fast. It started and stopped by a start-run capacitor. Newer units use a DC motor feed from a transformer and rectifier to allow soft start and battery backup.

Inside the opener’s AC motor there are two separate coils of wire wound around the stationary outer part, the stator. Each coil becomes an electromagnet when current flows through it.

Illustration 2 – The rotor always chases the field from the coil that peaks first toward the one that peaks second. Drag the slider to watch it, then switch the capacitor to the other coil and the rotor runs backward.

The capacitor matters because household AC power alone is a simple sine wave that would just make the motor buzz back and forth, not turn. The capacitor delays the current into one of the coils by a fraction of a cycle. So the two coils peak one after the other instead of together. Let’s name the coils Coil A and Coil B. That handoff between the coils makes the magnetic field sweep around the motor, and the rotor follows it. This is just like two hands spinning a wheel one after the other.

To reverse the garage door, switch the order of the capacitor firing: the capacitor on coil A and A lags, so the field sweeps one way. Put it on coil B and B lags, so it sweeps the other way. In many openers the two coils are identical, and the capacitor is wired between them. The “up” relay feeds power to one end and the “down” relay to the other, which decides which coil gets the delayed current.

Since the motor spins at about 1,500 RPM, a worm on the motor shaft drives a larger worm gear, which is often made out of nylon, which reduces from 10 to 40 to 1. A worm drive is good because it’s hard to back drive, which helps hold the door in place when the motor stops. This worm gear turns the output shaft, and the sprocket on that shaft pulls the chain. A pretty simple drivetrain.

Illustration 3 – Drag the slider to open the door. Drag the model to turn it, or tap Close-up to see the forces at the door arm.

The arrows in the illustration show the forces on the door at each point in its travel. The red arrow is gravity: the door’s 78 kg pulls down with about 765 N from its centre of mass. Gravity matters less as the door goes more horizontal. The magenta arrows are the lift from the torsion springs, delivered through the two cables. With the door closed they carry almost the whole 765 N, about 380 N per cable. As the door rolls onto the horizontal track, the track takes over more of its weight and the magenta arrows shrink to match. Because the springs balance the door, the opener only has to make up the difference. The dark cyan arrow is the opener’s force on the door, pushed through the door arm. At the start of opening it is only about 20 N, just enough to overcome roller friction and the drag of the bottom seal. The light cyan arrows split that force into a lift and a pull.

With the door closed the arm leans about 42° from vertical, so it pulls up and back at once, and the arm carries about one and a half times the chain’s pull. The springs do the heavy lifting; the motor steers and tops up.

The torsion spring is cool because it is not a normal spring with a linear constant that increases its force. It doesn’t pull on anything directly; it twists. One end is bolted to a bracket on the wall and cannot turn. The other end is clamped to the steel shaft that runs above the door. When the door is closed, the spring has been wound about seven turns.

Illustration 4 – Drag the slider to open the door. The wound spring twists the shaft, the drum reels in the cable, and the cable lifts the door. Watch the coils loosen as the spring gives up its twist.

Like a wound-up rubber band it is always trying to unwind, and the only part free to move is the shaft. So the spring twists the shaft, about 19 newton-metres from each spring in this example. At each end of the shaft is a small drum, about 10 cm across, with a steel cable wrapped around it and running down to the bottom corner of the door. As the shaft tries to turn, the drums try to reel those cables in. A twist of 19 N·m on a drum with a 5 cm radius becomes a pull of about 380 N on each cable, and two cables together roughly equal the door’s own weight (78 kg, or about 765 N).

The door is effectively hanging from a pair of wound-up springs, balanced. As the door rises, the cables wind onto the drums and the springs give up their twist. They lift less and less, which is fine, because the horizontal track is taking over more and more of the door’s weight. What a cool, balanced, system.

You can see a full simulation of all the forces at garagedoor.theboohers.org/ but you can get a small sample of everything together below. (Btw, perfect introduction to differential equations.)

Illustration 5 – Drag the door or press Open & close. Change the spring, door mass and friction to see how the forces and power change.

The Math

This is a great chance to show differential equations in action. The model has a single coordinate, $ s $: how far the bottom roller has travelled along the track. Every other position follows from it.

The whole door has one moving part as far as the math is concerned: $s$, how far the bottom roller has travelled along the track. Every section, roller, cable and the trolley can be worked out from $s$, so one equation describes the entire door.

In words: the force needed to speed the door up equals what the opener supplies, minus gravity, plus the spring, minus friction.

\begin{aligned} M(s)\,\ddot s + \tfrac12 M'(s)\,\dot s^{2} &= F_{\text{op}} \\ &\quad – G(s) \\ &\quad + S(s) \\ &\quad – F_f\,\operatorname{sgn}(\dot s) \end{aligned}

The left side is the door’s inertia: $M(s)$ is how heavy the door feels at position $s$, $\ddot s$ is its acceleration and $\dot s$ its speed. On the right, $F_{\text{op}}$ is the opener’s push, $G$ is gravity, $S$ is the spring’s lift and $F_f$ is friction, which always opposes the motion. The simulation runs this equation backward: it knows how the door moves and solves for $F_{\text{op}}$, the force the opener must supply at each point.

The door’s effective mass adds up every section’s sliding and turning:

\begin{aligned} M(s) &= \sum_{i=1}^{4} m_s \left\lVert \frac{d\mathbf c_i}{ds} \right\rVert^{2} \\ &\quad + \sum_{i=1}^{4} \frac{m_s h^{2}}{12} \left( \frac{d\phi_i}{ds} \right)^{2} \end{aligned}

Each of the four sections has mass $m_s$ and height $h$. The first sum is the sections sliding, with $\mathbf c_i$ the centre of section $i$. The second is the sections turning, with $\phi_i$ its tilt and $m_s h^{2}/12$ a flat panel’s resistance to rotating. Because the sections speed up and tilt as they round the curve, $M(s)$ changes along the track, and that is where the $\tfrac12 M'(s)\,\dot s^{2}$ term in the main equation comes from.

Gravity is the slope of the door’s potential energy:

\begin{aligned} U_g(s) &= \sum_{i=1}^{4} m_s\, g\, \bar y_i(s) \\ G(s) &= \frac{dU_g}{ds} \end{aligned}

$\bar y_i$ is the height of each section’s centre. $G$ is largest when the door is closed and hanging straight down, and falls toward zero as the sections lie flat on the horizontal track, where the track carries their weight.

The torsion spring unwinds as the cable winds onto the drum:

\begin{aligned} n(s) &= n_0 – \frac{L_0 – L(s)}{2\pi r_d} \\ T_c(s) &= \frac{k\, n(s)}{r_d} \\ S(s) &= -\,T_c(s)\, \frac{dL}{ds} \end{aligned}

The springs start wound $n_0$ turns, about 7 here. Every $2\pi r_d$ of cable wound onto a drum of radius $r_d$ gives back one turn, so $n(s)$ is the wind left. The spring’s twist is $k$ newton-metres per turn, and the drum turns it into the cable pull $T_c$. $S$ is that pull measured along the door’s travel. It is positive because the cable shortens as the door rises. A balanced spring makes $S$ track $G$ almost exactly.

Friction comes from the rollers pressing on the track:

\begin{aligned} N_j &= w_j\,\lvert t_z(r_j) \rvert + \frac{m_j v_j^{2}}{R} \\ F_f &= \mu \sum_{j} N_j \left| \frac{dr_j}{ds} \right| + F_{\text{seal}} \end{aligned}

$N_j$ is how hard roller $j$ presses on the track. On the vertical track $t_z = 0$ and the cables hold the door, so the rollers only guide it. On the horizontal track $t_z = 1$ and the full weight share $w_j$ rests on the roller. On the curve of radius $R$, a moving roller is also pushed outward by $m_j v_j^{2}/R$. Rolling friction is a small fraction $\mu$ (about 0.03) of that, plus a steady drag $F_{\text{seal}}$ from the bottom seal and hinges.

Finally, the opener and the electric bill:

\begin{aligned} F_{\text{trolley}} &= \frac{F_{\text{op}}}{\zeta'(s)} \\ P &= F_{\text{op}}\,\dot s = F_{\text{trolley}}\, v_t \\ P_{\text{wall}} &= P_{\text{standby}} + \frac{\max(P,\,0)}{\eta} \end{aligned}

The chain pulls a trolley, and the trolley pulls the door through the door arm. $\zeta'(s)$ is how far the trolley moves per unit of door travel, so where the trolley moves less than the door, the chain must pull harder. Power is force times speed, the same at the door or at the trolley (moving at $v_t$, about 0.18 m/s). The motor and gearbox are about 55% efficient ($\eta$), and the board and radio draw about 8 W on standby ($P_{\text{standby}}$). The $\max(P,\,0)$ means that when the door pushes back on the motor, that energy is lost as heat rather than returned to the wall.

The Math

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