# What diameter hole will a twist drill bit of a given size actually make in a piece of metal?

I have a piston of diameter 0.7 mm exactly and a 0.7 mm drill bit. Let's call the diameter of the piston $r_1$, the diameter of a hole at the end of a tube $r_0$, and the final-actual diameter of the tube hole $r$. Then with fluid being pushed by the piston through the hole at the end ($r_0$) I calculate that I need:

$$\pi (r^2 - r_1^2) \lt \pi r_0^2 \\ \iff r^2 \lt r_1^2 + r_0^2 \\ \iff r \lt \sqrt{r_1^2 + r_0^2}$$ in order to prevent back-flow in the negative direction of the piston movement. I'm ignoring gravity. With $r_0 =$0.05mm, $r_1 =$0.35mm, I come up with: $r \lt \sqrt{(0.05)^2 + (0.35)^2} \approx 0.3536$. So can I get a hole that's between 0.7 and 0.7072 mm (preferably closer to 0.7 mm) using a 0.7 mm bit?

It's okay if the tolerance is so tight that the piston gets stuck at room temp, because the piston is a rod of continually fed material that will be melted in the tube.

• I drilled a supposed 0.7mm hole, and it was too small. I guess I'll try 0.8 :) – StudySmarterNotHarder Apr 1 '16 at 2:17
• The best way I know of to make a hole with a precise diameter is to first drill the hole with a slightly smaller drill bit, then use a reamer (en.wikipedia.org/wiki/Reamer) to size the hole. Reamers are designed to be very accurate but can only remove a small amount of material from the bore. Also, have you considered temperature changes causing the bore to thermally expand when heated? – willpower2727 Apr 1 '16 at 11:52
• willpower2727 is correct. Start by drilling a smaller hole and then use a reamer to get your 'exact'(ish) size. Most of what we work with for press vs. slip fittings come in +/- .001" sizes and I bet it wouldn't be difficult to locate metric reamers with similar tolerances. – eatscrayons May 24 '16 at 17:23
• Along with overall size, the biggest problem with drilled holes is that they tend to not be round or straight – DLS3141 Jul 25 '17 at 21:46

I don't know about the math, but the answer to the stated question depends on how the bit is driven ( solid drill press , hand held , etc) and the type of drill/point. The point is generally not in the exact center so normally makes an oversize hole . To correct this, a "gun drill" only cuts on one side; and there are other types of points.

It looks like you're trying to use area to determine how much back-flow there will be. The flow will be depended on the pressure gradients, which are dependent not just on cross sectional area but also length.

It sounds like you're trying to model something like a 3D printer filament being pushed through a heated nozzle and want to determine how tight of a tolerance you'll need to prevent melted plastic from flowing back out the entrance.

This is a pretty complicated problem but I'll try to give the basics of it:

First we need to establish some pressures, and additional dimensions:

Now we can see the melted piston in green can flow in two directions back around the solid piston, or forward through the exit orifice. First let's look at the forward flow rate:

For fairly viscous media (like melted plastic) through thin pipes flow rate is calculated by the Hagen–Poiseuille equation:

$$Q= \frac{\pi\, R^4\,\Delta P}{8\,\mu\, L}$$

Where $Q$ is your volumetric flow rate, $L$ is the length of your tube ($L_0$), $R$ is your tube radius ($r_0$), $\mu$ is your viscosity (going to be highly temperature dependent), and $\Delta P$ is the difference in pressure across the tube ($P_1-P_2$)

So plugging in:

$$Q_0= \frac{\pi\, r_0^4\,(P_1-P_0)}{8\,\mu\, L_0}$$

Flow around a piston can be modeled like flow between two parallel plates. Except the plates are wrapped into a circle to form cylinders. When one plate is moving relative to the other this is called Couette Flow.

$$Q=\frac{\Delta P\,h^3 \, w }{12 \mu\,L} - \frac{U\,h\,w}2$$

Where $h$ is the height of our chanel $r-r_1$, $w$ is the width of our channel (in this case circumference of the piston) $2\pi r_1$, $U$ is the velocity of our piston, $\Delta P$ is $P_1-P_2$, and $L$ is $L_1$

Plugging in we get:

$$Q_1=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \pi\,U\,(r-r_1)\,r_1$$

Now we're looking for this second Q to be zero. This means all of the melted material gets pushed out the front hole. If the flow rate is positive, the edge of the melted plastic would flow further up the channel, increasing $L_1$ which would in turn decrease the flow rate. This is good news, it means as long as the sleeve is long enough the flow will stabilize around zero flow. Also since we are now assuming all the newly melted plastic is going to go out the exit, this allows us to relate the flow rate out the front hole with the velocity of the piston:

$$Q_0=U\, \pi {r_1}^2$$

Combining the equations:

$$0=Q_1=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \frac{Q_0\,(r-r_1)}{r_1}$$

$$0=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \frac{\pi\, r_0^4\,(P_1-P_0)\,(r-r_1)}{8\,\mu\,r_1 \, L_0}$$

Now, while viscosity will probably be different in the different regions due to temperature, it's super convenient if we ignore that because then we can just multiply it out.

$$\frac{(P_1-P_2)\,(r-r_1)^3 \, r_1 }{3\,L_1} = \frac{r_0^4\,(P_1-P_0)\,(r-r_1)}{4\,r_1 \, L_0}$$

Further rearranging:

$$4(P_1-P_2)\,(r-r_1)^2 \, {r_1}^2 \,L_0 = 3\,{r_0}^4\,(P_1-P_0)\,L_1$$

Similarly if ($P_1=P_2$) we can cancel that out:

$$4\,(r-r_1)^2 \, {r_1}^2 \,L_0 = 3\,{r_0}^4\,L_1$$

$$r=r_1+\frac{{r_0}^2}{r_1}\sqrt{\frac{3\,L_1}{4\,L_0}}$$

This will give you the maximum size bore you can use with the given lengths and other sizes of your hole.