# Medieval wooden screw press tonnage I need to calculate the tonnage a medieval wood screw press could generate. In particular I’m speaking of a press used to squeeze water from newly formed sheets paper in-between layers of felt (see picture) however medieval wood screw presses used to extract olive oil or juice from grapes are identical.

I am not an engineer but I will be giving a PechaKucha (a short 6 minute 40 second lecture) to a group of hand papermakers in a few months from now in which I will include the concepts of simple machines like the lever and the inclined plane to gain mechanical advantage.

Such presses exist in museums throughout Europe however being over 500 years old we are unable to use these to test the actual force. Here’s what I already know. The formula for calculating the ideal mechanical advantage on jack screws is straight forward which includes the advantage gained from length of the lever and the pitch of the thread. The much more difficult part is allowing for friction.

I’ve read that even well-lubricated jack screws have efficiencies of only 15% - 20% but that’s for a machined iron screw. The threads on these wooden screws were chiseled by hand with the most difficult part being the inside threads of the nut.

So my question is – is it possible to calculate the efficiency of a hand carved wooden screw? Presumably the upper limit of efficiency would be less than the 15% noted above (and is this 15% figure is correct in the first place)? But what might be the lower end of efficiency be? Here are the dimensions of the wood screw

• What other information can you give about the presses you are interested in? Any schematics or or names pictures of a specific screw I could look into? – MountainClimberi May 3 '17 at 19:03
• Sorry for taking so so long to respond, I do appreciate you taking the time to answer my question. Unfortunately I don't have any additional data to add and I wasn't sure exactly what you were asking. I can't do any type of testing on the museum screw and I don't have another wood screw I can test. I will add a drawing showing details of the thread however its does come down to the same question, the efficiency. – Brian May 25 '17 at 3:58
• What kind of efficiency are you looking for? – MountainClimberi May 25 '17 at 3:59
• There is no target efficiency. A friend is recreating 15th century paper and wants to use the same pressure as achieved by these medieval wood screw presses. I found a simple formula T.M.A. = 2 Pi r / Pitch. Using a 2500 mm long lever and a pitch of 58 mm I arrive at a figure of 271. If a man pushed on the lever with a force of 90 kg I arrive at 24,390 kg not allowing for friction. I’ve read that machined steel screw jack presses are 15 to 30% efficient, but what about a hand carved wood screw? I’m assuming that it would be at the lower end - 15% efficient or less? – Brian May 28 '17 at 15:30

## 2 Answers

I can only try to give you an approximate answer. The principle is simple. We give rotational energy $E_{\text{rot}}$ into the system and we get axial force $F_{\text{ax}}$ times displacement $\Delta x$, taking into account that the system is not ideal we also need to multiply the rotational energy by the efficiency $\eta$. Assuming the axial force is constant we can write:

$$\eta E_{\text{rot}}=F_{\text{ax}} \Delta x.$$

Now, for a screw with thread pitch $\tan(\alpha)$ and mean diameter $d$ the variable $\Delta x$ can be calculated for $N$ rotations as $\Delta x = N\pi d\tan(\alpha)$.

In order to determine $E_{\text{rot}}$ we assume a radial input force $F_1$ and a lever of length $r$. Hence, the rotational energy for $N$ rotations is given by

$$E_{\text{rot}}=N(2\pi r F_1).$$

Plugging these results into the first equation and solving for $F_{\text{ax}}$ yields:

$$F_{\text{ax}}=\eta \frac{E_{\text{rot}}}{\Delta x}=\eta \frac{N(2\pi r F_1)}{N\pi d \tan(\alpha)}=\eta\frac{2rF_1}{d\tan(\alpha)}.$$

So this system is nothing else than a gearbox. Depending on your input lever $r$ (can be measured or estimated from the picture), mean diameter of the screw $d$ (can be measured or estimated from the picture), the efficiency $\eta$ and the pitch $\tan(\alpha)$ (can be measured or estimated from the picture) it is possible to get almost every axial force that you like. The Force $F_1$ can also be estimated by pushing against a balance and measuring the weight. The problem is, that at some point the stresses inside the wooden screw will reach a limit and the wooden screw will begin to bulge. I don't know the exact value for which wood will start to bulge under compression, but this will dictate the practical limit for your axial force.

Estimation of efficiency

If you want to get the efficiency you simply solve the last equation for $\eta$. The other measures can be obtained by measurements. You need to do measurements as it is very unlikely that you will have an analytic solution for the efficiency without using any kind of measurements. One thing is for sure, the lower limit is an efficiency of zero.

$$\eta = d\frac{\tan(\alpha)}{2r}\frac{F_{\text{ax}}}{F_1}$$

The only unknown in the system is the axial force $F_{\text{ax}}$. Just take the screw and deform a metal with given Young's modulus $E$ (the law should almost be the same for compression and strain for "isotropic" materials). Measure the strain $\varepsilon$ and use Hook's law to get the stress $\sigma=E\varepsilon$. Then $F_{\text{ax}}=\sigma A_{\text{cross section}}.$

• I think a big part of the OP's problem is precisely in defining $\eta$. – Wasabi May 3 '17 at 19:47
• @Wasabi: Thank you for your comment I thought the force was important. I added an additional part. – MrYouMath May 3 '17 at 19:58
• @MrYouMath the question was "So my question is – is it possible to calculate the efficiency of a hand carved wooden screw?" how does your answer help? – Solar Mike May 4 '17 at 11:08

Lets say the worker (or pair of workers, front and back) that used this press could apply 100lbf or 445N of force. And lets say that force was applied at about 1 meter away from the center of the screw. The screw has a large force contact area and would be well coated in viscous lard for lubrication. The lubrication would be in the "fluid film" regime in which the fluid supports all the load; the same regime a modern grease lubricated lead screw works in. Modern lead screws are in the 30 to 50% efficiency range. Page 9 of this paper calculated 25% efficiency with a friction of 0.2 and an angle of 30 degrees. Lets say 10% efficient to be conservative.

Force * Distance = 2*Pi* efficiency * torque

Force = (2* 3.14159 * 0.10 * (445N * 1m)) / 0.0584m
Force = 4788N or 1076lbf (per screw)

The area it presses is important too. Just guessing based on your photo I'm going to estimate 25cm x 25cm. This gives a pressure of 77kPa or 11psi. This must have been sufficient pressure to economically press the paper; otherwise the medieval engineer/artisan would have used a smaller pressing area to get a higher pressing force.

More pressure would result in more water removed, but it is not linear. The increase from 0 to 5 psi will remove more water than the increase from 5psi to 10psi, and so on. So even if the press generated more or less pressure, its not super critical from a historical process point of view. The operator was probably most concerned with the paper holding shape after it was pressed. The raw materials in the paper, pressing aids (gypsum), upstream processes, etc will also have a big factor on water retention/removal.

Pressed moisture would have likely still been 60 to 75% (still lots of water to remove). Using the press would have saved a lot of time and fuel for the a secondary drying process. Sun or fire would have been necessary to get the moisture level down to a non-molding level (say 5 to 10%). (My experience in this area is from beet pulp pressing and drying, but those numbers should be close.)