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Metric or traditional inch system?

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I am not sure about the eastern United States. But at least west of the Mississippi, legal land descriptions use a base line and meridian method. (Think x-and-y axes.) The units used are miles and acres. The land is broken into townships emanating from the intersection of the base line and meridian. Each township is six miles by six miles, giving rise to thirty-six sections within each township. Since a section (square mile) is 640 acres, land descriptions are something like

The north 1/2 of the east 1/2, Section 16, Township 6N, Range 14E Mt. Diablo Baseline and Meridian. This would be the legal description of the quarter section (160 acres) in Section 16, the sixth township north and fourteenth range east using the summit of Mt. Diablo as the intersection of the x-and-y axes. Sections within each township are numbered left to right starting in the upper left corner, then continuing right to left so that Section 7 is immediately beneath Section 6, and Section 12 is immediately beneath Section 1. Then Section 13 is beneath 12 and so on.

It is conceivable that the measurements of smaller, irregularly-shaped lots that are surveyed could easily be converted from feet to meters, and the angles converted from degrees to radians. But I think it would be nearly impossible to restate the legal descriptions. Not to mention the headaches in the recorder's office and the assessor's office.

And don't forget most survey work is also done in rods/chain (360 and 80 chains/mile, 4 rods per chain) - remember that an acre is 1 chain by 1 furlong (and 1 furlong is 10 chains) - so - we get into a decimal system (like metric) quickly - an acre is 10 square chains. A mile is 8 furlongs

Those "odd" units become 'Not so odd' when you realize they are a nice consistent measurement system when used with one another
The so called "Imperial" measures aren't really one measurement system, but at least 3 - the inch/foot/yard, the survey unit system, and the nautical system, each optimized for the job it was intended to do. BTW, part of the reason for the 8/80 etc (and why 60 for time) - How many factors are there of the number 10? The number 60? The number 80?
 
Billions upon billions upon billions of dollars. Probably trillions. if you think it'd be cheap, you aren't thinking hard enough.

All the signs… twice (you think it can be done without a transitional period? Think again). Huge changes to legacy software. Retraining a hell of a lot of people that rely on the system not changing. Surveyors?. Car nav systems (and speedometers in some cars). Textbooks. Thermometers. Stoves. Thermostats. Blueprints. So so so so much more. LAWS! Think of how many laws are explicitly worded using imperial. Then you've gotta actually educate people on this stuff. How long do you think it'll take Americans to get used to thinking their body temperature should be ~37 instead of ~98.6? The aviation world at large mostly uses nautical miles and feet. Should they change too?

On top of that, all the cultural stuff like hand-egg and baseball.

of all those examples all I agree on is that the longest is a cultural difficulty spanning several generations. The rest would be changed within one generation. Is that a long time? Yes. But it’s not a reason to not do it.
 
We were told we were switching over completely in the 1970's and it was mandatory to teach the metric system in schools. We ended up with some cars that were 1/2 metric and 1/2 imperial, and never did complete the changeover.
Thank you Jimmy Carter for letting this go :-(

Gerald Ford started metrification in 1975, The metrification board was abolished in 1982 by President Ronald Reagan, thanks Ronnie!
Actually, it was more Carter than Reagan. Carter let it die, Reagan just signed the death certificate.

Carter is an awesome ex-president, but his term in office was horrible.

But that's kind of expensive to buy 2 sets of tools.
Most (US) mechanics already have two sets of tools: Imperial and Metric

How about temperature. 100 degrees to boiling and 0 to freezing.
Alas, that is only at sea level. The boiling temperature drops as the altitude climbs. For example, at 1,905 meters (6,250 ft - about 1.2 miles), it is 93.4 C (200.1 F). At the top of Mt. Everest, you would only have to get to about 68 C (155 F) to boil. Granted, other factors weigh in, but I';m not smart enough to know what that might be.

However, the tires industry still use 'inches' to measure wheel diameters, but millimeters for the width!
There was a brief stint (in the US) of metric diameter tires. My Dad had a Thunderchicken Thunderbird (early 80's I believe) that used them. Ugly-ass car - looked like someone took an El Camino style car, but took the front of the hood from another car, turned it around and stuck it on the back.

Right, well, how many centimeters is 20,000 leagues?
You're in deep water at that point.
 
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Right, well, how many centimeters is 20,000 leagues?
20,000 leagues = 1.1112e+10 cm - just work out 3 nautical miles to the league, go from there - and the nautical mile is a VERY logical unit - 1 minute of arc latitude. IF they had gotten the meter right (it was supposed to be 1/10000000 the distance from the equator to the pole at Paris - aka 5400 nautical miles - it was supposed to be even, but the guys doing the inventing of the meter got the diameter wrong
 
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Alas, that is only at sea level. The boiling temperature drops as the altitude climbs.

Actually water has not boiled at 100 degs C (at STP) for a while now, it is actually 99.974 degrees, as ITS-90 went and redefined the 0 pints using the triple point of water (and 13 other substances), and then THAT changed again last year
  • 2019 definition: The kelvin, symbol K, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be 1.380649×10−23 when expressed in the unit J⋅K−1, which is equal to kg⋅m2⋅s−2⋅K−1, where the kilogram, metre and second are defined in terms of h, c and ΔνCs.
The kelvin may be expressed directly in terms of the defining constants as:

1 K = 1.380649×10−23/(6.62607015×10−34)(9192631770)hΔνCs/k = (2+032485708115445338/121822045942277331)hΔνCs/k.