How to recreate units of measure?

Question

How is it possible to recreate standard units of measure (meter, kilogram, second) with Renaissance-level technology?

A man (let's call him "Traveler") gets transported to a different world, which is very much Earth-like. This Traveler was able to bring a good amount of knowledge from modern day Earth, and this can be very useful because this world's tech level is barely Renaissance. However, Traveler could not bring any material object with him, not even the simplest tape measure. Naturally, this creates problems in recreating newer technology, because all of the knowledge that Traveler has is based on metric system.

Most of all, Traveler is bothered by the gravitational acceleration of this new world. If feels "off" to him, not by much, but he is pretty sure that it can be up to 10% different from Earth's standard 9.81 m2/s.

Would it be possible to accurate measure gravity of the new world? Traveler knows his height, and this gets him within 1% accuracy of the meter, but has no reference for kilogram or second? Traveler can talk to the brightest minds of this new world, and they are open to the idea of conducting experiments, however there are not yet any Galileos among them.

Related question: How to create precise measurements from scratch deals with the same problem, however, in that question starting conditions, as well as criteria of success were different.

P.S. A clarification - Traveler had an opportunity to prepare for this trip and memorize various blueprints and reference materials. His plan was to recreate SI units once he arrived, but different gravity was an unexpected factor which stopped him from being able to accurately recreate kilogram and second.

Answer 1

To measure gravity accurately, you need to be able to measure both distance and time accurately. These two are fundamental measures of existence. The other fundamental unit that cannot be measured without reference but only defined is mass.
(yes, you could quickly count off 6.0221409e+23 atoms of carbon, and know that's 12.0107grams. But it might take you a while)

Almost everything else that exists are merely derivatives of these three things.

Your traveler has a moderately accurate scale for distance (his own height, although if gravity is different that's not so very accurate, his spine will compress or elongate differently!! Also remember you are about 1% longer when you get up from sleep, than you were when you got in bed), and will have to derive a suitable scale for time.

Unfortunately, just about the only absolute constant (assuming all laws of physics cooperate) would be the speed of light. He would need to remember the speed of light, measure the local speed of light in arbitrary time units + his known length , and from that derive an accurate measure of time.

I suspect this will take a while, and a lot of effort.

Some quicker approximation, and suitable for the tech level:
Use the speed of sound. He is breathing normally, so can assume atmospheric conditions and compositions very similar to Earth. Using his length, measure off 171.5m from a handy cliff. The time between him making a noise and the echo returning is one second.

Answer 2

You've got it backwards.

He doesn't need to recreate the SI units precisely, at least not until the science and technology is sufficiently advanced.

What he needs is a stick of roughly one meter, and a weight of roughly one kilogram, and a timer of roughly one second.

He probably knows his arm span, as this is used routinely to make quick and dirty measurements. If he doesn't know it, then he probably knows that for normally proportioned men, the arm span is about the same as the height. (See the Vitruvian man.) This will give him a meter with about 5% accuracy.

^{Yes, the question says that he knows his height to the centimeter. Unfortunately, this is just a convention. The height of a person varies by more than 1% from waking up in the morning (when the spine is relaxed) to going to bed in the evening (when the spine is compressed from supporting the weight all day long). Add in a slightly different gravitational acceleration...}

A kilogram, long time ago, was defined as the mass of a cubic decimeter of water. The modern definition is very much fancier, but it comes to very nearly the same thing. So he has a decent approximation of a kilogram, say ±15%.

He probably knows the rate of his pulse rate at rest. If he doesn't, he can take 70 beats per minute, if he is an ordinary man, or 55 beats per minute, if he is an athlete. (If his pulse rate is much different from those values, he will definitely know it.)

This will give him a second with ±10% precision.

So he has a meterr which is ±5% of an Earth meter, a kilogramm which is ±15% of an Earth kilogram, and a sekond which is ±10% of an Earth second. Notably, his kilogram has the same relationship to the meter as the Earth kilogram does, which means that all his knowledge about densities of materials still holds.

This is plenty good enough for practical work. His knowledge won't be exact, but it will be close enough.

He can measure the gravitational acceleration in meterrs per sekond squared by measuring the weight of a kilogramm; or he can measure it by measuring the length of a pendulum with beats sekonds. Of course, this won't give the gravitational acceleration in Earth newtons, but this would be a useless piece of data, with zero practical importance.

Now comes the hard part: those new units of measurement have to be used for all engineering work. But, fortunately, the question does ask about this.

Time passes. Centuries pass, and the society develops science and engineering roughly equivalent to the 1960s or 1970s on Earth. At this point, if they are curious, they can easily determine the relationship of their meterr, kilogramm and sekond to Earth's meter, kilogram, and second, using the ordinary definitions of the Earth units as they were in that timeframe.

Answer 3

Problem extension: He doesn’t really have to recrate the same units.

I’ll explain: In physics, the only thing that is really tied to the units is the constants (they will change values if you change units).

The formulae themselves won’t change. $\sum \vec{F} = m \vec{a}$ will still be true, in your new units of force, mass, speed etc. (as long as your unit system is coherent).

But you suggested constants (though not fundamental ones) like the gravity of the planet might be different. That means he’ll have to recompute those constants anyway if he wants to use them.

So why bother trying to reconstruct the SI system? Might as well define new units that are easy to compute on that planet (distance relative to the diameter of the planet, temperature relative to the freezing/boiling points of water on that planet, etc.), and recompute the constants in that unit system.

Answer 4

If he knows his height, he can get a reasonable approximation of a meter length. If he knows different ways of counting of seconds ("one one thousand, two one thousand"), he can get a reasonable approximation of a second.

Now all he needs is a string and a rock.

Tie one end of a string to a branch, the other to the rock, and make sure the total length is the meter. Then let it gently swing, and count the seconds it takes to make 10 complete swings (start-out-back). At 1g, a meter-long pendulum has a period of 2.0064 seconds, so 10 swings should take 20 seconds. If it takes less than 20 seconds, the gravity field is stronger. If it takes longer, less gravity.

It won't be much (19 vs 20 vs 21 seconds) for a 10% difference in g, and honestly given the approximations for the measurement of length and time may be drowned out, but it's something. But if it seems to consistently be different in several tests, combined with his feeling of things being heavier or lighter, it could be enough to confirm. And you don't need mass at all to make the determination.

Answer 5

Length: Measure your height in advance and use it to approximate a centimeter.

Time: Measure your resting BPM in advance and use it, along with a clock to approximate a second.

Gravity: Use a long pendulum with length $L$ and a small starting angle and the formula $$T = 2 \pi \sqrt{\frac{L}{g}}$$ to compute the force of gravity. The formula simplifies to get $$g = \frac{4 \pi^2 L}{T^2}.$$

You can compute the value of pi experimentally by drawing circles or by winding a string around a circular pole.

Mass: One kilo is the mass of a 10cm cube of water. You already know the centimetre so this is easy.

Weight: On Earth one kilo weighs 9.8 Newtons. On the new planet where gravity has strength $g$ the cube weighs $g$ Newtons. So on the new planet a cube with 9.8/g times the volume will weigh 9.8 newtons or the same as the original cube on Earth. Such a cube has side $10 \sqrt[3]{9.8/g}$ centimetres.