#14r

[sfmuniphotos]

14R Mission Rapid

[arinewman7]

Conception of Alexander the Great

Bruges, ca. 1468-1475

British Library, Burney 169, fol. 14r

[motorcycle-id]

2012-2022 Kawasaki Ninja ZX-14R

they put hte mmster cneif in th mobercyrcle

[teamassociated]

New! Parts for the Reflex 14R Hoonitruck RTRs

These are the new parts for the Reflex 14R Hoonitruck and LiPo Combo RTRs.

Available: April 2023. (Pictured: #20177 and 20177C Reflex 14R Hoonitruck and LiPo Combo RTRs.)

More: https://tinyurl.com/asc20177parts

[mylaptopspares]

Buy Lapgrade Battery for Dell Inspiron 13R 14R 15R 17R

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[cuties-in-codices]

mary as the "woman clothed with the sun"

in a manuscript containing alchemical illustrations, southern germany, c. 1570-1580

source: Vatican, Bibl. Apostolica Vaticana, Pal. lat. 1885, fol. 14r

[the-mediaeval-monk]

Bodleian Library MS. Bodl. 270b f.14r

The citizens of Sodom surrounded by demons

[nessvn]

One of my favourite miniatures from BnF MS fr606, "L'Epistre d'Othea la deesse, que elle envoya a Hector de Troye, quant il estoit en l'aage de quinze ans" by Christine de Pizan. 15th century.

The image depicts Apollo and Pan playing the harp and the flute for Midas to judge. Folio 14r.

[nowimjustastranger]

Gravity Falls Mini-Series Masterlist

We'll Be Better Tomorrow

Ford 419"3 from the STCMO series pays Stan Pines from Dimension 90V'1 a visit with two very familiar toddlers who need a home. Or, the Dad Stan AU

I Promise I'll Do Better

Ford 419"3 from the StCMO series delivers a baby Stan Pines from Dimension J0\65 to old man Ford Pines doorstep in Dimension Xi11-5 with the intention of having old man Ford raise baby Stan. Or, the Dad Ford AU

Stronger Together

In Dimension 83;9F, Stan and Ford went through the portal together, traveling the multiverse together. They make a living as bounty hunters, no longer focusing on getting back to their home dimension. Or, the Feral Ford AU

So Far From Home

In Dimension 38^Pi, Stan Pines wipes Gravity Falls off the map with a glorified bomb after his brother was captured and the niblings were killed. StCMO Ford saves him, bringing him to Dimension 14R-1, where Stan Pines killed himself after tricking Bill into possessing him. Or, Stan is relocated after nuking his dimension.

[felixcloud6288]

Okay. I really want to talk about that second math problem Homura did. It's the only one where I was able to understand what the actual problem she's trying to solve was (Her mathwork in q1 and q3 were correct for what I saw, but I couldn't figure out what she was ultimately trying to do.)

The problem: We're given three integers x, a, and b where a/14 has a remainder of 6, b/14 has a remainder of 1, and x^2 - 2ax + b = 0. Find the remainder of x/14.

Step 1: We define x, a, and b in relation to 14. We'll use q, s, and t to represent the quotients of x, a, and b respectively and r to represent the unknown remainder of x/14.

x = 14q + r a = 14s + 6 b = 14t + 1

Step 2: We rewrite x^2 - 2ax + b = 0 using the above equalities.

(14q + r)^2 - 2(14s + 6)(14q+r) + (14t + 1) = 0

Step 3: rewrite each term using FOIL (Normally I'd say simplify but there's nothing simple about what this is going to look like. Also, I'm going to keep 14 as unedited as possible for reasons you'll see later).

(14q)^2 + 2*14qr + r^2 - 2*14^2qs - 2*14rs - 12*14q - 12r +14t + 1 = 0

Step 4: Our end goal is to figure out what happened when x is divided by 14. So we're going to collect all terms divisible by 14 together and factor out 14.

14(14q^2 + 2qr - 28qs - 2rs - 12q + t) + r^2 - 12r + 1 = 0

Step 5: This was something Homura did but didn't explicitly show. Consider this: -14r = -12r - 2r. So we could add 2r - 2r into the left side of the equation without impacting it. This let's us throw -14r into the pile of terms being multiplied by 14 and leave us with a 2r we can use FOIL on with the remaining terms

14(14q^2 + 2qr - 28qs - 2rs - 12q + t) + r^2 - 12r + 1 + 2r - 2r= 0 14(14q^2 + 2qr - 28qs - 2rs - 12q + t) + r^2 - 14r + 1 + 2r = 0 14(14q^2 + 2qr - 28qs - 2rs - 12q + t - r) + (r+1)^2 = 0

Step 6: This is the part where Homura went from equations to logic. To simplify things, let's let y = 14q^2 + 2qr - 28qs - 2rs - 12q + t - r. This means:

14y + (r+1)^2 = 0 (r+1)^2 = -14y

Now I have to go over a little bit of number theory. Prime factorization is the process of breaking a positive integer down into a series of prime numbers. With the above equation, we can say (r+1)^2 is divisible by 14 since it is equal to 14 times some unknown integer -y. This also means (r+1)^2 can be divided by 2 and 7 since 2*7=14.

Now consider: Let's say we have two integers m and n. There is a unique set of prime numbers that can be multiplied to get m and a unique set that can be multiplied to get n. If we multiply m by n, the prime factors of mn is going to be the prime factors of m and n. For example, 12 = 2*2*3 and 15 = 3*5 so 12*15 = 2*2*3*3*5 or 2^2 * 3^2 * 5.

Now let's say we multiplied m by m. In that case, we're combining the same prime factors together. So m^2 would double the exponent values of each prime factor. As a corollary, if m's square root is an integer, then m's prime factors must all be exponentiated by an even number.

Now let's tie this back to the problem.

Currently, we know (r+1)^2 = -14y for some unknown integer y. Our end goal is to figure out r, the remainder when x is divided by 14. First, we need to try getting rid of the square on the left side. Going back to what I mentioned earlier though, since r+1 is an integer, that means it has a certain prime factorization. We can say that prime factorization includes at least a 2 and 7 because of the 14 on the right. And since r+1 is being squared, that means (r+1)^2 has the same prime factors as r+1 but the exponential values are double. In other words, y has to also be divisible by 14 in order for this equation to be true.

Now also recall that r is the remainder of x/14. This would mean r at most is 13. So r+1 is at most 14. If we take all this information into account, this means (r+1)^2 can at most be 14^2 and in fact would have to be 14^2 because it is the only number that would work with all the previous information (Meaning that y = -14). Therefore:

(r+1)^2 = 14^2 r+1 = 14 r = 13

back

[insidemyfavoritemanuscript]

Episode 9: Zoe and Mac from The Maniculum Podcast!

In Episode 9 of Inside My Favorite Manuscript, Lindsey and Dot chat with Zoe and Mac from the Maniculum podcast, where they suggest ways to adapt medieval texts for TTRPG’s (tabletop roleplaying games). We talk about marginalia, games, and Mac takes us for a dive into the Rutland Psalter. We also talk to Zoe about her storytelling for Pentiment, a medieval adventure video game by Obsidian Entertainment.

Listen here, or wherever you find your podcasts.

If you enjoyed this episode, you can listen to Dot and Lindsey over on The Maniculum Podcast! We talked about digital humanities, book history, and the secrets of women. Find links to listen here: https://www.themaniculumpodcast.com/episodes

Below the cut are page images and further reading from our conversation.

The Maniculum Podcast

The Maniculum Podcast on Tumblr

Pentiment

Book Historian Plays Pentiment! (Allie Alvis Pentiment play through)

Ludohistory Pentiment Playthrough Playlist on YouTube

Perna (oyster) - Der naturen bloeme - Jacob van Maerlant - KB KA 16 - 108v (via Wikimedia Commons)

Sebhat the Ethiopian scribe relates the story of Lazarus in Pentiment (via Obsidian on Twitter)

Saint George and the Dragon, UPenn LJS 102 f. 71r, early 20th century Ethiopian manuscript

British Library Add MS 62925, The Rutland Psalter

f. 14r

Close-up of bottom of 14r

Close-up of top of 14r

f. 57r

Close-up of bottom of f. 57r

f. 88v

Close-up of bottom of 88v

Betsy Chunko-Dominguez (2016) “Playing on Timbrels”: the margins of the Rutland Psalter, Word & Image, 32:1, 131-141, DOI: 10.1080/02666286.2016.1146540 (PDF here)

f. 67r

Close-up of bottom (heh) of f. 67r

f. 66v

Close-up of bottom of f. 66v

How the opening of ff. 66v-67r would approximately appear

The Luttrell Psalter on the BL Website

The Luttrell Psalter film, mentioned at this point in the podcast, is embedded at the end of this post.

Susan Kim and Asa Mittman, Inconceivable Beasts: The ‘Wonders of the East’ in the Beowulf Manuscript. ACMRS, 2013. (Download PDF here)

British Library blog post on Cotton Tiberius B v

Maniculum D23 posts

f. 87v

Close-up of bottom (heh) of f. 87v

f. 64r

Close-up of bottom of f. 64r

Close-up of top of f. 64r

The Book of Kells

f. 88r

Close-up of bottom of f. 88r

f. 16r

Close-up of bottom (heh) of 16r

(I promise I’m not doing this on purpose)

[Images in the 170s]

f. 5v, Calendar page for October

Close up of Scorpio zodiac sign

f. 3v, Calendar page for June

Close-up of Cancer Zodiac sign

f. 78v

Close-up of f. 78v.

f. 70v

Close-up of f. 70v

Further reading on The Rutland Psalter:

Chunko-Dominguez, Betsy. “‘Playing on Timbrels’: The Margins of the Rutland Psalter.” Word & Image (London. 1985), vol. 32, no. 1, 2016, pp. 131–41 (PDF)

Morgan, Nigel. “The Artists of the Rutland Psalter.” The British Library Journal, vol. 13, no. 2, 1987, pp. 159-185 (PDF)

Luttrell Psalter Film on YouTube:

[sfmuniphotos]

A 14R-Mission Rapid bus traveling rapidly through the Mission District

[myloverlist]

2023 F1 14R - Dutch Grand Frix | Carlos Sainz

[favemusiclessons]

Paranoid, if it were written by Dire Straits

Listen to the full 'if it were written by' series on Spotify: https://open.spotify.com/playlist/14R... I thought it was about time to dive into the heavier stuff, so I went with a Black Sabbath song. I had to bring the tempo down a bit, from 164 bpm to 151, to fit the style a bit better. Even then it's still a tiny bit faster than Sultans of Swing or Lady Writer, which are both at 148 bpm. I made the choruses a little more mellow to add some dynamics and of course topped it off with a solo at the end.

[usafphantom2]

6 January 1949. First flight of the prototype Nord 1400-01 Noroit F-WFDL. French reconnaissance and air-sea rescue flying-boat. Cantilever gullwing monoplane with a two-step hull. Powered by two 1,600 hp Gnome-Rhone 14R star engines.

@ron_eisele via X

[bobby69lund]

My 2013 Kawasaki ZX 14R

We delve into the realm of speed and precision with the legendary Kawasaki ZX-14R (2013). Known as a superbike with unparalleled power and cutting-edge technology, the ZX-14R continues to be a force to be reckoned with. Join us as we take a closer look at the Kawasaki ZX-14R (2013) and discover what makes it an extraordinary machine. Dominating Powerhouse:At the heart of the Kawasaki ZX-14R…