# How long is the 85th parallel? **URL:** \ **Category:** Factual Questions\ **Created:** [February 12, 2014, 3:38am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117 "2014-02-12T03:38:04Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Joe\_King](https://avatars.discourse-cdn.com/v4/letter/j/fbc32d/32.png) [@Joe\_King](https://boards.straightdope.com/u/Joe_King)\ **Post date:** [February 12, 2014, 3:38am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/1 "2014-02-12T03:38:04Z")
I know that the equated is ~ 24,000 miles but having a hard time figuring out how many miles each degree increases the distance. I know someone can help me out. ---
**Author:** ![OldGuy](https://avatars.discourse-cdn.com/v4/letter/o/3bc359/32.png) [@OldGuy](https://boards.straightdope.com/u/OldGuy)\ **Post date:** [February 12, 2014, 3:45am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/2 "2014-02-12T03:45:07Z")
> [@Joe\_King](#): > > I know that the equated is ~ 24,000 miles but having a hard time figuring out how many miles each degree increases the distance. > I know someone can help me out. Assuming a spherical earth, any latitude should be 24,000\*cos(latitude) = 2092 miles ---
**Author:** ![Svejk\_1](https://avatars.discourse-cdn.com/v4/letter/s/ea666f/32.png) [@Svejk\_1](https://boards.straightdope.com/u/Svejk_1)\ **Post date:** [February 12, 2014, 3:49am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/3 "2014-02-12T03:49:16Z")
When you hand in your homework, make sure you spell ‘equator’ right 🆒 ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [February 12, 2014, 4:20am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/4 "2014-02-12T04:20:10Z")
> [@Joe\_King](#): > > I know that the equated is ~ 24,000 miles It’s actually a tad over 24,900 miles. ---
**Author:** ![davida03801](https://avatars.discourse-cdn.com/v4/letter/d/b5e925/32.png) [@davida03801](https://boards.straightdope.com/u/davida03801)\ **Post date:** [February 12, 2014, 2:42pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/5 "2014-02-12T14:42:57Z")
Btw……… the distance \*\*decreases \*\*as you move north or south from the Equator. ---
**Author:** ![Northern\_Piper](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/northern_piper/32/5304_2.png) [@Northern\_Piper](https://boards.straightdope.com/u/Northern_Piper)\ **Post date:** [February 12, 2014, 6:19pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/6 "2014-02-12T18:19:08Z")
> [@OldGuy](#): > > Assuming a spherical earth, any latitude should be 24,000\*cos(latitude) = 2092 miles Doesn’t the length of a circle of latitude decrease the closer you go to the North or South Poles? ---
**Author:** ![Joe\_King](https://avatars.discourse-cdn.com/v4/letter/j/fbc32d/32.png) [@Joe\_King](https://boards.straightdope.com/u/Joe_King)\ **Post date:** [February 12, 2014, 6:26pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/7 "2014-02-12T18:26:14Z")
> [@Northern\_Piper](#): > > Doesn’t the length of a circle of latitude decrease the closer you go to the North or South Poles? I think this is why I am having a hard time finding the answer. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [February 12, 2014, 6:34pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/8 "2014-02-12T18:34:01Z")
> [@Northern\_Piper](#): > > Doesn’t the length of a circle of latitude decrease the closer you go to the North or South Poles? I’m not sure I understand your question, but on the face of it, yes the latitude lines are smaller the closer you get to the poles, and that’s why Northern Piper included the “cos(latitude)” term in the formula. ETA: Nevermind, I think it was a joke that whooshed me at first. ---
**Author:** ![buddha\_david](https://avatars.discourse-cdn.com/v4/letter/b/ee7513/32.png) [@buddha\_david](https://boards.straightdope.com/u/buddha_david)\ **Post date:** [February 12, 2014, 7:12pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/9 "2014-02-12T19:12:33Z")
The 85th parallel is much shorter than 24,000 miles. It’s only 5 degrees away from either Pole, where the latitudes become more circular than linear. [Here’s a map of Antarctica](http://geology.com/world/antarctica-satellite-image.shtml) to demonstrate the concept – note the smallest latitude shown (80°) is well within the boundaries of the Southern Continent. ---
**Author:** ![bump](https://avatars.discourse-cdn.com/v4/letter/b/7c8e57/32.png) [@bump](https://boards.straightdope.com/u/bump)\ **Post date:** [February 12, 2014, 7:24pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/10 "2014-02-12T19:24:45Z")
> [@OldGuy](#): > > Assuming a spherical earth, any latitude should be 24,000\*cos(latitude) = 2092 miles Here’s how I thought about it: The earth’s circumference at the equator is 24901 miles, which is a full 360 degrees. From the equator to the north pole/south pole is only 90 degrees, so 1/4(24901) = 6225 miles. 1/90th of that is 69.169 miles per degree of latitude. 69.169 \* 5 - 345.85 miles. That’s the radius of a 5 degree circle centered around the pole. 2 \*π \* r is the formula for circumference. So you get 2(3.14159)(345.85) = 2173 miles as the circumference of the circle. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [February 12, 2014, 7:49pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/11 "2014-02-12T19:49:18Z")
> [@bump](#): > > Here’s how I thought about it: > > The earth’s circumference at the equator is 24901 miles, which is a full 360 degrees. > > From the equator to the north pole/south pole is only 90 degrees, so 1/4(24901) = 6225 miles. > > 1/90th of that is 69.169 miles per degree of latitude. > > 69.169 \* 5 - 345.85 miles. That’s the radius of a 5 degree circle centered around the pole. > > 2 \*π \* r is the formula for circumference. So you get 2(3.14159)(345.85) = 2173 miles as the circumference of the circle. Using your numbers since I can’t be arsed to look up the actual difference in polar and equatorial circumference. Your method only works for a circular flat Earth with a radius of 6225 miles, and for the first few degrees of latitude south of the north pole that works (or north of the south). But at 45 degrees of latitude your method gives a circumference of 19557, while the real number is 24901\*cos(45) = 17607.7 miles. And your “equator” is 39114 miles. Maybe you intended to think in a way appropriate for an approximation near the poles, but that’s not clear from your post. ---
**Author:** ![Pleonast](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/pleonast/32/1183_2.png) [@Pleonast](https://boards.straightdope.com/u/Pleonast)\ **Post date:** [February 12, 2014, 7:53pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/12 "2014-02-12T19:53:01Z")
> [@bump](#): > > Here’s how I thought about it: > > The earth’s circumference at the equator is 24901 miles, which is a full 360 degrees. > > From the equator to the north pole/south pole is only 90 degrees, so 1/4(24901) = 6225 miles. > > 1/90th of that is 69.169 miles per degree of latitude. > > 69.169 \* 5 - 345.85 miles. That’s the radius of a 5 degree circle centered around the pole. > > 2 \*π \* r is the formula for circumference. So you get 2(3.14159)(345.85) = 2173 miles as the circumference of the circle. Of course, the Earth is more of an oblate spheroid than a sphere. That is, the radius of the Earth is slightly smaller going from the center to the North pole than from the center to the Equator. Using a slightly better model of the Earth’s shape, I get the length of the 85th parallel to be 2163 miles. Given that that’s less than 0.5% different, I don’t think we need to worry about the precise shape. ---
**Author:** ![johnspartan](https://avatars.discourse-cdn.com/v4/letter/j/a183cd/32.png) [@johnspartan](https://boards.straightdope.com/u/johnspartan)\ **Post date:** [February 12, 2014, 8:24pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/13 "2014-02-12T20:24:03Z")
> [@Xema](#): > > It’s actually a tad over 24,900 miles. Is that a tad in space terms? [![](https://img.youtube.com/vi/Gd6aLnPHqeE/maxresdefault.jpg "Airplane 2: The Sequel (5/10) Movie CLIP - Don't Panic (1982) HD") ](https://www.youtube.com/watch?v=Gd6aLnPHqeE) ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [February 12, 2014, 9:23pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/14 "2014-02-12T21:23:42Z")
Just to be clear on a point above. For a sphere the answer is cos(lattitude) \* equatorial circumference. Where **bump** ’s linear approximation works is for angles close to 90˚. This is always worth remembering, cos(ø) ≈ ø for for values of ø close to 90˚, and sin(ø) ≈ ø for values of ø close to 0. For a much more accurate approximation to the Earth, we can note that the radius of the Earth at the poles is less than that at the Equator. At the poles the radius is 6,356.75km. So a line of latitude at 85˚ has a radius of 6,356.75 \* cos(85˚) = 554.03km and the circumference is 3481km or 2163 miles. ie, the answer **Pleonast** provides. ---
**Author:** ![OldGuy](https://avatars.discourse-cdn.com/v4/letter/o/3bc359/32.png) [@OldGuy](https://boards.straightdope.com/u/OldGuy)\ **Post date:** [February 12, 2014, 9:51pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/15 "2014-02-12T21:51:05Z")
> [@Francis\_Vaughan](#): > > This is always worth remembering, cos(ø) ≈ ø for for values of ø close to 90˚, and sin(ø) ≈ ø for values of ø close to 0. Well it would be a bit better to remember cos(pi/2-ø) ≈ ø for for values of ø close to 0. Since both those formulas only work if you measure angles in radians and you want the complement for cosines. ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [February 12, 2014, 10:05pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/16 "2014-02-12T22:05:54Z")
> [@OldGuy](#): > > Well it would be a bit better to remember cos(pi/2-ø) ≈ ø for for values of ø close to 0. Since both those formulas only work if you measure angles in radians and you want the complement for cosines. Oops, quite right. I was trying to avoid radians, as most people are used to degrees, and sort of forgot I had done so. Exactly, the approximation only works in radians. :smack: ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [February 12, 2014, 10:56pm UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/17 "2014-02-12T22:56:44Z")
> [@Francis\_Vaughan](#): > > For a sphere the answer is cos(lattitude) \* equatorial circumference. > Where **bump** ’s linear approximation works is for angles close to 90˚. This is always worth remembering, cos(ø) ≈ ø for for values of ø close to 90˚, and sin(ø) ≈ ø for values of ø close to 0. Yes, but if you’re making approximations like these, you don’t state your answer to five significant figures like bump did. I’m all for making good approximations, but please don’t overstate the accuracy. ---
**Author:** ![jtur88](https://avatars.discourse-cdn.com/v4/letter/j/e9c0ed/32.png) [@jtur88](https://boards.straightdope.com/u/jtur88)\ **Post date:** [February 13, 2014, 12:26am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/18 "2014-02-13T00:26:21Z")
The 85th parallel is the circumference of a circle whose radius is 5/360 of the circumference of the earth. About 3,500 km. That would be off by a very small allowance for the curvature of the radius line from the pole to the parallel , almost negligible at that latitude. No trigonometry required. ---
**Author:** ![bump](https://avatars.discourse-cdn.com/v4/letter/b/7c8e57/32.png) [@bump](https://boards.straightdope.com/u/bump)\ **Post date:** [February 13, 2014, 4:44am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/19 "2014-02-13T04:44:35Z")
> [@Francis\_Vaughan](#): > > Just to be clear on a point above. > > For a sphere the answer is cos(lattitude) \* equatorial circumference. > Where **bump** ’s linear approximation works is for angles close to 90˚. This is always worth remembering, cos(ø) ≈ ø for for values of ø close to 90˚, and sin(ø) ≈ ø for values of ø close to 0. > > For a much more accurate approximation to the Earth, we can note that the radius of the Earth at the poles is less than that at the Equator. At the poles the radius is 6,356.75km. So a line of latitude at 85˚ has a radius of 6,356.75 \* cos(85˚) = 554.03km and the circumference is 3481km or 2163 miles. ie, the answer **Pleonast** provides. You’re really quibbling over 10 miles difference? Sheesh. ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [February 13, 2014, 5:08am UTC](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117/20 "2014-02-13T05:08:17Z")
> [@bump](#): > > You’re really quibbling over 10 miles difference? Sheesh. Of course, this is the dope, we worry things to death. 😃 Actually, the approximation I used isn’t correct either. Whilst the distance from the pole to the centre of the earth is less than the distance of centre to the equator, the radius of curvatures are the other way around - the earth curves more at the equator than at the pole. So the poles are flatter. The radius of curvature at the pole is larger than the radius of curvature at the equator, and thus the 85th parallel is actually longer than that on a perfect sphere, not less. At 85˚ it seems to be about 6398km. So the answer is 3503.6 km or 2177 miles. Still only 5 miles out from the back of the envelope calculation - which shows how useful such simple approximations are. [Next page](https://boards.straightdope.com/t/how-long-is-the-85th-parallel/681117.md?page=2)