Hi, my name is Becca and I am a senior in highschool with a math puzzel of sorts. I dreamed this up one day and I have consulted many people on this and every one of them has given me a different answer. My question is this: If it takes two nails to secure a plank from rotating on a wall in the third spatial dimension, how many nails would it take if you were attempting to secure the board from rotating in the fourth spatial dimension? Friction between plank and nail doesn't come into play in this scenario. Even if this question has no answer, I would like the input of someone who knows about this sort of thing. My email address is becca_bober@hotmail.com. I would really appreciate a reply. Thank you for your time.


Hi Becca,

If you are securing a board to a wall (with nails) then you really are REDUCING the problem joining two 2-dimensional surfaces - the wall surface and the wall side of the board.

When ONE nail is in, you are left with only one degree of freedom - the rotation about the nail. The nail itself is, effectively, giving a 3-D axis for a single motion of the plank (relative to the wall). It is a 'thick' (rotationally symmetric) piece of a line - ie. a cylinder.

For 4-space, there would be several answers - depending on whether you are NOW using '4-D' nails or '3-D' nails. A 4-D nail would leave just one degree of freedom between the plank and the wall. It would be like the 4-D axis for a single rotation - i.e. a thick (rotationally symmetric) piece of a plane - form of cylinder in 4-space.

With this in place, you just need one more attachment to hold the plank still - could be another 4-D nail, OR a 3-D nail, or even a 2-D nail (usually called a ball joint in 3-space). All of these are redudant (do more than necessary).

IF you are using 2-D nails (ball joints) in 3-space. Then it is possible that you actually need THREE such nails to hold two 3-D objects together. (Sure, with a flat wall and a flat board you can TRY to use the friction and pressure of flatness to hold them from the other rotations after you install a ball joint. However, if you put TWO ball joints between the wall and the board, along one edge, it would still have a rotation about the line through the two ball joints.)

Now consider using 3-D nails in 4-space. One such nail would leave several degrees of rotation. In a cross section (a 3-D space perpendicular to the line of the nail) it would look and feel like a universal ball joint in 3-space. An additional 3-D nail inside THIS space, would still hold the plank to the board. [The two 3-D nails in 3-space did MORE than was necessary - and they still are enough in 4-space.]

There ARE general counting principles for this type of problem, in any dimension. The KEY is to break down the effect of something complicated, like a nail, into a set of individual constraints or linear equations. I will give a brief outline for the way a structural engineer might count things - using statics and local counts.

In the plane, each 'body' has 3 degrees of local freedom

  • 2 directions of translation
  • 1 of rotation around the point
All other motions are (instantaneously) a linear combination of these.

To hold two full 2-D bodies together, we need to remove all three of these possible motions between them. Therefore we need THREE constraints. However a 2-D nail is 2 constraints - so 2 x 2 (2-D nails) = 4 which is more than enough.

In 3-D, a full 'body' has 6 degrees of freedom

  • 3 directions of translation
  • 3 axes of rotation
All motions are a combination of these.

To hold two 3-D objects together you need to remove all 6 motions between them. We need SIX constraints. However, a 3-D nail is FIVE constraints (leaves only one motion if used correctly). 5 x 2 (two nails) = 10 > 6 so two nails will do.

IF you use 2-D nails (ball joints) each of these is THREE constraints. It would appear that 3 x 2 = 6 is enough BUT the two nails SHARE a constraint so two 2-D nails actually end up like a 3-D nail - removing 5 dgrees of freedom. [If you know matrices and linear algebra, I am saying that the two sets of three equations share a common equation, and a row reduction of the matrices would give a zero row etc.]

In 4-D, a body has 10 degrees of freedom

  • 4 translations
  • 6 rotations.
You need 10 constraints to hold two bodies together.

A 4-D nail would remove 9 degrees of freedom. A 3-D nail could remove 5 degrees of freedom (maybe up to 7 depending on how it was made). A 2-D nail (a point of attachment) would probably remove only 4.

If I am getting a bit less precise here, it is because there IS ambiguity about exactly WHAT a nail is when manufactured in 4-D! That ambiguity probably explains why you get different answers. There ARE assumptions being built in at each stage of the discussion.

One way out is to concentrate on the 'single' constraints: what an engineer would call a 'bar' with 'universal joints' on the two ends. Then the count is clear, and we only need to make sure the 'bars' are spread out well. IN the plane, you join two bodies by three bars - making sure the three bars do NOT pass through a point. In 3-D you join two bodies by six bars, making sure the six lines are independent (not easy to explain that unless you know projective geometry - line complexes. These days, most engineers don't know the precise way to say that - though everyone knows ONE way to fail have all six lines of the bars intersect a single line.)

The SAME problem comes up in reverse in ROBOTICS. How many pistons do you need to move an aircraft simulator around to simulate full 3-D motions? 6 (in something called the Stewart Platform). ALSO how many hinge joints to you need in a robot arm to ensure that the fingers of the hand can take all positions and orientations in a region. Six. (They often use seven to give easier planning.)

In 4-D the count says 10 bars properly spread out. In 5-D the count says 15 bars. For N-D the count is (N+1)(N)/2 bars. Check that this works for the previous cases.

It was a GREAT QUESTION - and leads lots of different directions! Ambiguity is, in fact, normal in these kinds of investigations - meaning you still have choices left to make. Keeps life interesting and IS part of applied mathematics at all levels.

Walter Whiteley
Math and Stats, York University

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