So there's an easy rule of thumb that people sometimes use, where you square the number of miles and multiply that by 8 inches.  While some have pooh-poohed that, I've seen charts to where it's accurate to a few feet out to over 100 miles, but didn't see it farther than that.

Using that formula for 175 miles yields 20,416 feet, whereas the real trigonometric calculation is somewhere in the 19,000+ range.  Not enough to make a difference regarding the point being made here, and it's easy for anyone to scribble down or even do in their head.

Now, one does have to take into account the elevation of the observer, which the FE folks in these videos always do, since it's critical.  Given the elevation of the obsever, in the video, of 1,000 feet, that results in 12,000+ feet of hidden height, and the elevation of the highest peak of the Canagu range is only 9,000.  That would mean the entire then should be hidden with about 3,000 feet to spare, and yet we see that most of it is visible, with the exception of maybe 1,000 - 2,000 at the bottom due to various factors.  I don't care how you crunch numbers, unless the globe is 10x bigger than the claim, there's no way that physically only the bottom portion of the range would be obscured, and you'd need some massive refraction to pull it off.

So, another reason that Canigou is critical as an FE proof ... it happens every year.  Refraction is inherently inconsistent, where some years it wouldn't happen at all, other years it would be badly distorted, etc. etc. ... and yet this same exact phenomenon is consistenlty see every year.

https://earthcurvature.com/

https://dizzib.github.io/earth/curve-calc/index.html?d0=175&h0=10&unit=imperial

So, the first calculator you provided, earthcurve.com, is correct and does work and proves my calculation as well as the calculator I'd used. The problem is that you used the wrong distance: the distance in earthcurve.com is from the observer to the halfway point to the object (mountain) in the distance, or in this case, 87.5 miles. The curvature of the Earth (cord height) between two points is measured halfway between them. At 87.5 miles, the result is nearly the same as in my first post:
That other calculator you sent makes no sense and apparently does not apply to what we're talking about. Look at the results from entering 175 miles and 0 ft. observer elevation. It comes back with "0" for d1. What does that even mean  ?:

The distance is the full amount.  There’s nothing on that site that says to use 1/2 the distance. 

No, and the 8 inches times miles squared gets you very close to the same numbers.  I'm sure that if these calculators were doubling the distance they would have been called out long ago by the Globers.  I can do the math again, but I have done it for degrees of latitude and longitude for various parts of the alleged globe.  It's accurate.

That sounds like the voice of Jeran, one of the flat earthers who went to Antarctica and, for the record, he is now a glober so it would be interesting to hear his thoughts on this now.

I don't know. I think I remember reading about that about a year ago, and it is not conclusive. It seems like an Ockham's Razor kind of deal. There would need to be more thorough travel and exploration to prove anything. I mean if the earth is a globe, they should try to fly across Antarctica in four evenly intersecting pathways and witness the alleged South Pole.