What if the Earth's axis were tilted 90° to the ecliptic?

Answer 1

Axis becomes equator. Equator becomes Axis.

Answer 2

The ecliptic would be perpendicular to the celestial equator (from your question).

On the day of the "Spring" equinox. The Sun's declination is zero and moving northward (i.e., on the next day, the Sun appears north of the celestial equator).

Three months later (solstice), the Sun's declination is 90˚ N (The Sun appears directly overhead at the North pole). On that day, all locations with northern latitude have the Sun over the horizon all day (no night). For all locations south of the equator, the Sun does not climb above the horizon (no daylight).

Actually, because of refraction and the diameter of the Sun's disk, you'd still have daylight (a 24-hour-long "sunset") down to latitude 1.5˚S, a 24-hour long civil twilight down to 7.5˚S and a 24-hour-long Nautical twilight down to 13.5˚S. In that last interval, the sky is dark enough to see the bright stars, yet light enough for the horizon to still be visible (it is possible to use a sextant to find a ship's position using nautical astronomy, hence the name "Nautical" twilight).

Another three months and it's Autumn equinox (the beginning of Autumn in the Northern hemisphere, the beginning of Spring in the Southern hemisphere).

And yet another three months takes us to winter solstice (the Sun appears directly overhead at the South pole).

On the celestial sphere, the Sun's average apparent movement (along the ecliptic) would be a little bit less than 1 degree per day.

However, I do not know how long the days would be (probably around 22 of our present hours). The Moon's orbit is presently closer to the plane of the ecliptic than the Earth's equatorial plane: 5 degrees from the ecliptic (our equatorial plane is 23.4˚ from the ecliptic). I suspect that the Moon of your proposed Earth would be a bit further from the ecliptic (pulled a bit by Earth's equatorial bulge, but the Sun's influence would still be the major influence).

Because of the high inclination of the Moon's orbit (compared to the equator), the tidal braking effect on Earth's rotation would not have been as pronounced. Therefore, I believe that the Moon would appear a bit larger than it does now (longer total eclipses of the Sun) and Earth's days would be a bit shorter.

The Sun's Right Ascension would remain (almost) the same for 6 months at a time. At Spring equinox, the Sun's R.A. would be 0 (same as the present definition).

Then, until solstice, it stays on the same R.A. and only the declination increases. Lines of R.A. are perpendicular to the celestial equator, and so is the ecliptic (by your command).

When the declination reaches 90˚, the R.A. flips by 180˚ (or 12 hours, as R.A. are counted in "hours" where 1 hour = 15 degrees) and the Sun's declination decreases to reach 0˚ on Autumn equinox; it continues south for another 3 months on the same R.A. until the declination reaches 90˚S

Then it "flips" again and the Sun climbs the line of 0 R.A. until its declination reaches 0˚ again, after one full year.

Because of the shorter days, there would be more days in a year. If we are lucky, your perpendicular Earth has 400 days in a year. This means that we would use grades (100 in a right angle) instead of degree (90˚ in a right angle).

Units like 360, 60, 24 and 12 were chosen because of the way they can be divided by all kinds of integers (like 2, 3, 6) AND because 360 was very close to the number of days in a year (the Sun moving "almost" one degree per day).

So, with 400 days, we'd pick angle units so that 400 represent a complete circle (and that is what grades do). The grade would be divided in 100 centigrades. Then, our lenght unit would definitely be the kilometer (it represents a distance subtended by an angle of 1/100 grade at the centre of the Earth -- the nautical mile is the distance subtended by 1/60 of a degree). There would be 100 days in each season (of course, the eccentricity of Earth's orbit would make it so that some have a few more days, others have a few less).

Keeping with the "hundred" divisions, units of time could be chosen so that a day lasts 10 units, each unit divided in hundreds. So, the time of day would be given in the form: 3:82:25 (3 "hours", 82 "minutes" and 25 "seconds" where each hour has 100 minutes and each minute has 100 seconds). The new second would be about 0.8 of our present seconds.

Also, with a Moon closer to us, we may have more lunar months in a year (let us say a little over 13.8 compared to the present 12.4). Maybe we'd have months of 30 days and years of 13-13-14 months (a three year cycle). Three years = 40 months of 30 days = 3 times 400 days.

Presently, on Earth, the "tropics" represent the region where the Sun can be seen directly overhead at least once per year (dotted lines at 23˚26.7' N and S latitudes on the globe). The arctic circle (and the antarctic circle in the South) shows the southernmost latitude at which it is possible to have the sun above the horizon at midnight.

On your perpendicular Earth, the entire planet would be the tropics; the equator would be the arctic and antarctic circle(s).

Assuming a circular orbit around the Sun, the Sun's declination increases by 1 grade per day from -100 at winter solstice to +100 at summer solstice.

Halfway into Spring (50 days after the equinox), the Sun's declination is 50 grades (45˚). Someone who lives at a latitude of 45 grades (40.5˚, e.g., New York) would have the sun above the horizon for 330/400 of the day = 8.25 of the 10 "hours"

Using:
Length of daylight= 2P, where P is found with
Cos(P) = - tan(LAT)*tan(DEC) = tan(50)*tan(45) = 165 grades
2P = 330 grades
10 "hours" = 400 grades
therefore 330 grades = 330/400 times 10 = 8.25 hours.
Don't forget to use grades (not degrees) and the new "hours" (10 longer hours in a shorter day)

The length of day increases rapidly: 5 days later, the Sun's declination reaches 55 grades and the Sun no longer sets for New York, until its declination comes back down to 55 grades (90 days later).

On Spring equinox, days and nights are equal in New York (and everywhere else that is not within 300 km of the poles).

Days after equinox -- Sun above horizon

000 -- 5 "hours" (half the day)
010 -- 5.43
020 -- 5.90
030 -- 6.43
040 -- 7.13
050 -- 8.25
060 -- 10 (all day)
070 -- 10
080 -- 10
090 -- 10
100 -- 10 (solstice)
110 -- 10
120 -- 10
130 -- 10
140 -- 10
150 -- 8.25 (finally, dark nights)
160 -- 7.13
170 -- 6.43
180 -- 5.90
190 -- 5.43
200 -- 5 (Autumn equinox)
210 -- 4.57
220 -- 4.10
230 -- 3.57
240 -- 2.87
250 -- 1.75
260 -- 0 (the long night)
270 -- 0
280 -- 0
290 -- 0
300 -- 0 (winter solstice)
310 -- 0
320 -- 0
330 -- 0
340 -- 0 (Sun briefly seen from tallest buildings?)
350 -- 1.75
360 -- 2.87
370 -- 3.57
380 -- 4.10
390 -- 4.57
400 -- 5 (Happy New Year)

(The symmetry is because I assumed a circular orbit for Earth).

As for the weather patterns...