# Optimal door opening

This is a problem that has been periodically bugging me, so I finally decided to work on it. I haven't done any physics since high school, so I'm a bit out of practice:

Consider a doorway with two hinged doors, one being the mirror of the other. Someone opens one of the doors (which open outward) and lets go, allowing the door to begin shutting. You want to walk though the doorway, but you need one of the doors to be fully open. When is it more optimal to open the closed door completely, as opposed to pulling the closing door back open? By optimal, I refer to the torque over time I would have to exert to open the door fully in time $t_{open}$

Specifications

• The doors use 2 manual door closers (a spring that is stretched when the door is open and pulls the door back) which are connected the top and bottom of the door (for simplicity) in such a way that they always pull perpendicular to the door frame
• The doors have two sets of hinges equidistant from the center of the door (again, for simplicity)
• The doors are fully open when they reach an angle $\theta_{open} < {\frac{\pi}{2}}$ from the frame.

Assumptions:

• Ignoring friction/air resistance/gravity entirely
• ignoring dampening mechanisms of door closer (I have no idea how they work)
• Treating the door as though it had no width

Attempt

So, we have a door of mass $m$ with length across $l_d$, and a handle offset some distance $l_h$ from the hinges of the door. There is are two springs connected to the inner side of the door some distance $l_{s}$ from the hinge edge.

Let the two springs each have a spring constant $\frac{k}{2}$. When offset some distance $x$, they pull on the door with force of magnitude $kx$, giving a torque on the door of $l_{s}kx\cos{\theta}$. The offset $x$ is related to $\theta$ by $l_{d}\sin{\theta} = x$, so torque is equal to $l_{d}l_{s}k\sin{\theta}\cos{\theta}$, and torque over time is equivalent. Since we can treat the door as a thin rod, moment of inertia $I$ of the door as $\frac{ml_{d}^{2}}{3}$, we know the angular acceleration $\alpha$ of the door: $$\alpha(t) = \frac{3l_{s}k\sin{\theta}\cos{\theta}}{ml_{d}}$$ Then, the angular velocity $\omega$ is : $$\omega(t) = \frac{3l_{s}k\sin{\theta}\cos{\theta}}{ml_{d}}t$$

and $\theta$ is: $$\theta(t) = \frac{3}{2}\frac{l_{s}k\sin{\theta}\cos{\theta}}{ml_{d}}t^2$$

At this point, I pretty sure I've taken a wrong turn somewhere, probably involving conversion of angular acceleration to velocity, since theta is a function of time as well...

I feel like this is a problem I should be able to do relatively simply. Anyone want to help me along? Also, it seems like I'm probably over-complicating the situation...

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This seems to me to be a very complicated problem. Not that the physics is hard, but there are so many variables. When you swing open a door the work you do on the door goes into three things:

1. Elastic energy stored in the spring
2. Kinetic energy of the door
3. heating the fluid in the viscous damper

Plus, depending on how hard you push the door will dissipate different amounts of energy when it hits the doorstop. Unless you want to do a complicated calculation including all these variables you either need to ignore some of them, which is unphysical, or go out and measure them for a real door and specify them as constants in your calculation.

I think I see what you're getting at: If the door is swinging back towards you it's harder to open because you have to first stop it then put in the effort to open it. On the other hand if it's swinging away from you it's easier to push because it's already moving in the direction you want it to go. So somewhere in between the effort will be exactly the same as required to open a static door.

I would start with an oversimplified version, play with that and then look at making it more realistic. For example I'd take the torque due to the door spring as constant, so the work done in opening the door is just the torque $\tau$ times the angle $\theta$ ($\theta = \pi$/2 when fully opened). So opening a static door requires an energy input of $\tau\pi/2$. Suppose the door has swung closed by an angle $\theta$, its kinetic energy will be $\tau\theta$ so you need to put in energy $\tau\theta$ to bring it to a halt, then another $\tau\theta$ to push it back to fully open - so $2\tau\theta$ in total. For this to be equal to the energy needed to open the static door you just have:

$$\tau\frac{\pi}{2} = 2\tau\theta$$

so:

$$\theta = \frac{\pi}{4}$$

If the open door has swung back by less that 45 degrees it's easier to push at the closing door, while if it has swung back by more than 45 degrees it's easier to push the static door on the other side.

As the door swings through the closed position and back towards you it starts slowing again, and using the above method you can show that for $\theta$ > 135 degrees it once again becomes easier to push the swinging door. So the only time it's better to push on the static door is when the opened door is swinging towards you and $\theta$ is between 45 and 135 degrees.

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