I wish you could upload another video explaining in a little more depth how the parallax angle is obtained..... I'm still unsure how to calculate it from the apparent displacement of the nearby star
@@astropictionary3988 Hey work on it for me too? I was unsure about triangle name of how I fail to completely do it. But I have not understood it differently, anyhow probably not applicable like yours. A 180^ triangle equals 1/2 of a 360^ square. Any known angle having a known opposite side length becomes a ratio scale for that triangle. Whereby any other known angle may be multiplied to obtain its unknown opposite side's length. And any known side length can be divided by same ratio to find its opposite angle. True? I am trying to remember what should reasonably make sense, while trying not to remember someone measured Earth's land circumference "metrically" which gives me a metricked up headache because 40,000 something kilometres is some fraction 1/4 or 1/8 of planet's circumference. I attempt not to care before I am (((cussing))). But perhaps if if if an entire day was measured by 40 metronomes¿ somethings or different number of different somethings, I cussing already willing to bow to whatever "anti-familiarity political SUPPRESSOR's confusion" declares. I don't care whether switching away from base-10 arithmetic: actually that is another Suppressor trick example, because 9 is highest single digits number unless the zero is meant to represent an invisible fruitioned duplicate? Even when applying base-7, value eight is represented by number 10. I don't care how frustratingly complicated math is like everything is stupidly keeping soldiers and everyone else closer to verge of violence. As I understand it for myself: planet Earth has a radius of 4,000 miles and if a meteor or asteroid is not calculated to swerve at least 4,500 miles away, then someone is gambling the farm. Whatever Earth's Astronomical Unit probably the Sun's center is the other end of it. I continue guessing two sightings of objects opposite sides are taken at 6 o'clock mornng and another pair of sightings at 6 o'clock evening which diameter equals an 8,000 mile long triangulation side. All 4 angles at the Sun's diameter vertices are ratioed to Earth's 8,000 mile diameter side but the Sun's angles will vary according to what Astronomical Units of whatever exact distances, upon those seasonal days. Those very same 4 angles are also verticed opposite direction towards distant star. Okay probably the longer end of the rectangles are where I don't understand a sine from a co-sine norva stop sign. So how does it work from there without beginning World War 3? Maybe something different than my previous experiences in-completed by non-explainers thus far? Your turn, please?
How do we know the nearby and the distant stars are not moving, and how do we know that the background is so far away that it can be an used for this measurement?
Stars do move (see e.g., proper motion at ua-cam.com/video/p-ho_ttkFfE/v-deo.html) but the further they are, the more motionless they appear to move. Those that do move, move very little in the sky over the time frame we are watching a nearby object moving and this very little motion of stars we call the background.
This is a vocabulary word in textbook Astronomy Essentials by Michele M. Montgomery, Ph.D that contains Part 1 - Background and Basics on Apple Books: books.apple.com/us/book/id1590873139, Part 2 - Solar and Exosolar Systems on on Apple Books: books.apple.com/us/book/id1590877272, Part 3 - Sun, Space Weather, and Formation on Apple Books: books.apple.com/us/book/id1590879246, Part 4 - Stars and Stellar Evolution on Apple Books: books.apple.com/us/book/id1590881365, Part 5 - End Products and Binary Star Evolution on Apple Books: books.apple.com/us/book/id1590884825, Part 6 - Galaxies on Apple Books: books.apple.com/us/book/id1590887070
Hey work on it for me too? I was unsure about triangle name of how I fail to completely do it. But I have not understood it differently, anyhow probably not applicable like yours. A 180^ triangle equals 1/2 of a 360^ square. Any known angle having a known opposite side length becomes a ratio scale for that triangle. Whereby any other known angle may be multiplied to obtain its unknown opposite side's length. And any known side length can be divided by same ratio to find its opposite angle. True? I am trying to remember what should reasonably make sense, while trying not to remember someone measured Earth's land circumference "metrically" which gives me a metricked up headache because 40,000 something kilometres is some fraction 1/4 or 1/8 of planet's circumference. I attempt not to care before I am (((cussing))). But perhaps if if if an entire day was measured by 40 metronomes¿ somethings or different number of different somethings, I cussing already willing to bow to whatever "anti-familiarity political SUPPRESSOR's confusion" declares. I don't care whether switching away from base-10 arithmetic: actually that is another Suppressor trick example, because 9 is highest single digits number unless the zero is meant to represent an invisible fruitioned duplicate? Even when applying base-7, value eight is represented by number 10. I don't care how frustratingly complicated math is like everything is stupidly keeping soldiers and everyone else closer to verge of violence. As I understand it for myself: planet Earth has a radius of 4,000 miles and if a meteor or asteroid is not calculated to swerve at least 4,500 miles away, then someone is gambling the farm. Whatever Earth's Astronomical Unit probably the Sun's center is the other end of it. I continue guessing two sightings of objects opposite sides are taken at 6 o'clock mornng and another pair of sightings at 6 o'clock evening which diameter equals an 8,000 mile long triangulation side. All 4 angles at the Sun's diameter vertices are ratioed to Earth's 8,000 mile diameter side but the Sun's angles will vary according to what Astronomical Units of whatever exact distances, upon those seasonal days. Those very same 4 angles are also verticed opposite direction towards distant star. Okay probably the longer end of the rectangles are where I don't understand a sine from a co-sine norva stop sign. So how does it work from there without beginning World War 3? Maybe something different than my previous experiences in-completed by non-explainers thus far? Your turn, please?
@@samuelhmullins2170 It sounds like you are looking for the "law of sines" which is sinA/a = sin B/b = sin C/c for a triangle where A, B, C are angles and a, b, c are opposite side lengths to the angles. Look up "law of sines". As for the Earth problem, it looks like you are looking for an s=r x theta equation where r is the distance to the asteroid, s is the path the asteroid made in 12 hours , and theta is the angle Earth rotated in 12 hours. Since Earth rotated 12 hours and 24 hours is one rotation or 360 degrees, Earth rotated 180 degrees in 12 hours. If you know either s or r, then you can find the other. Because of the curvature of the Earth, measurements will likely need to be made at times less than 12 hours, say 8 am and 6 pm. For the last part, look up trigonometry of sine angle and cosine angle so you can see how sin (angle) = opposite/hypotenuse, which is hard to explain without imagery.
@@astropictionary3988 My lifetime's STUDENT-handicap is only slightly MORE advantaged than Nig-awanity ( all religiosities, self-governmental agencies incapatibilities' becoming unblinded compatible :: beyond nowaday's veil predestinated into non-gression, progressively foundationed all futuristics) . Thank you for gracious hope advancing my adjustments to avail potential's purposed for more of your remedial professorship. ). Pool-hall village's jargon, we will LEAVE OURSELVES OUT FROM BEHIND THE TABLE'S 8-BALL.. ( -er )....... then, not whenever and not if ever!!! Thanks for participating FAITH within tangibility optional God's Family Angels. Love you amidst my predicament regardless, of impediments today.
I wish you could upload another video explaining in a little more depth how the parallax angle is obtained..... I'm still unsure how to calculate it from the apparent displacement of the nearby star
let me work on this for you
@@astropictionary3988 Hey work on it for me too? I was unsure about triangle name of how I fail to completely do it. But I have not understood it differently, anyhow probably not applicable like yours. A 180^ triangle equals 1/2 of a 360^ square.
Any known angle having a known opposite side length becomes a ratio scale for that triangle. Whereby any other known angle may be multiplied to obtain its unknown opposite side's length. And any known side length can be divided by same ratio to find its opposite angle. True?
I am trying to remember what should reasonably make sense, while trying not to remember someone measured Earth's land circumference "metrically" which gives me a metricked up headache because 40,000 something kilometres is some fraction 1/4 or 1/8 of planet's circumference. I attempt not to care before I am (((cussing))). But perhaps if if if an entire day was measured by 40 metronomes¿ somethings or different number of different somethings, I cussing already willing to bow to whatever "anti-familiarity political SUPPRESSOR's confusion" declares.
I don't care whether switching away from base-10 arithmetic: actually that is another Suppressor trick example, because 9 is highest single digits number unless the zero is meant to represent an invisible fruitioned duplicate? Even when applying base-7, value eight is represented by number 10. I don't care how frustratingly complicated math is like everything is stupidly keeping soldiers and everyone else closer to verge of violence.
As I understand it for myself: planet Earth has a radius of 4,000 miles and if a meteor or asteroid is not calculated to swerve at least 4,500 miles away, then someone is gambling the farm. Whatever Earth's Astronomical Unit probably the Sun's center is the other end of it. I continue guessing two sightings of objects opposite sides are taken at 6 o'clock mornng and another pair of sightings at 6 o'clock evening which diameter equals an 8,000 mile long triangulation side. All 4 angles at the Sun's diameter vertices are ratioed to Earth's 8,000 mile diameter side but the Sun's angles will vary according to what Astronomical Units of whatever exact distances, upon those seasonal days. Those very same 4 angles are also verticed opposite direction towards distant star.
Okay probably the longer end of the rectangles are where I don't understand a sine from a co-sine norva stop sign. So how does it work from there without beginning World War 3? Maybe something different than my previous experiences in-completed by non-explainers thus far? Your turn, please?
OMG THANK YOU SO MUCH I LITERALLY HAVE A TEST TOMORROW AND MY TEACHER'S DISCUSSION WAS SO UNCLEAR
You are so welcome! I hope you did well!!
How do we know the nearby and the distant stars are not moving, and how do we know that the background is so far away that it can be an used for this measurement?
Stars do move (see e.g., proper motion at ua-cam.com/video/p-ho_ttkFfE/v-deo.html) but the further they are, the more motionless they appear to move. Those that do move, move very little in the sky over the time frame we are watching a nearby object moving and this very little motion of stars we call the background.
This is a vocabulary word in textbook Astronomy Essentials by Michele M. Montgomery, Ph.D that contains
Part 1 - Background and Basics on Apple Books: books.apple.com/us/book/id1590873139,
Part 2 - Solar and Exosolar Systems on on Apple Books: books.apple.com/us/book/id1590877272,
Part 3 - Sun, Space Weather, and Formation on Apple Books: books.apple.com/us/book/id1590879246,
Part 4 - Stars and Stellar Evolution on Apple Books: books.apple.com/us/book/id1590881365,
Part 5 - End Products and Binary Star Evolution on Apple Books: books.apple.com/us/book/id1590884825,
Part 6 - Galaxies on Apple Books: books.apple.com/us/book/id1590887070
Hey work on it for me too? I was unsure about triangle name of how I fail to completely do it. But I have not understood it differently, anyhow probably not applicable like yours. A 180^ triangle equals 1/2 of a 360^ square.
Any known angle having a known opposite side length becomes a ratio scale for that triangle. Whereby any other known angle may be multiplied to obtain its unknown opposite side's length. And any known side length can be divided by same ratio to find its opposite angle. True?
I am trying to remember what should reasonably make sense, while trying not to remember someone measured Earth's land circumference "metrically" which gives me a metricked up headache because 40,000 something kilometres is some fraction 1/4 or 1/8 of planet's circumference. I attempt not to care before I am (((cussing))). But perhaps if if if an entire day was measured by 40 metronomes¿ somethings or different number of different somethings, I cussing already willing to bow to whatever "anti-familiarity political SUPPRESSOR's confusion" declares.
I don't care whether switching away from base-10 arithmetic: actually that is another Suppressor trick example, because 9 is highest single digits number unless the zero is meant to represent an invisible fruitioned duplicate? Even when applying base-7, value eight is represented by number 10. I don't care how frustratingly complicated math is like everything is stupidly keeping soldiers and everyone else closer to verge of violence.
As I understand it for myself: planet Earth has a radius of 4,000 miles and if a meteor or asteroid is not calculated to swerve at least 4,500 miles away, then someone is gambling the farm. Whatever Earth's Astronomical Unit probably the Sun's center is the other end of it. I continue guessing two sightings of objects opposite sides are taken at 6 o'clock mornng and another pair of sightings at 6 o'clock evening which diameter equals an 8,000 mile long triangulation side. All 4 angles at the Sun's diameter vertices are ratioed to Earth's 8,000 mile diameter side but the Sun's angles will vary according to what Astronomical Units of whatever exact distances, upon those seasonal days. Those very same 4 angles are also verticed opposite direction towards distant star.
Okay probably the longer end of the rectangles are where I don't understand a sine from a co-sine norva stop sign. So how does it work from there without beginning World War 3? Maybe something different than my previous experiences in-completed by non-explainers thus far? Your turn, please?
@@samuelhmullins2170 It sounds like you are looking for the "law of sines" which is sinA/a = sin B/b = sin C/c for a triangle where A, B, C are angles and a, b, c are opposite side lengths to the angles. Look up "law of sines". As for the Earth problem, it looks like you are looking for an s=r x theta equation where r is the distance to the asteroid, s is the path the asteroid made in 12 hours , and theta is the angle Earth rotated in 12 hours. Since Earth rotated 12 hours and 24 hours is one rotation or 360 degrees, Earth rotated 180 degrees in 12 hours. If you know either s or r, then you can find the other. Because of the curvature of the Earth, measurements will likely need to be made at times less than 12 hours, say 8 am and 6 pm. For the last part, look up trigonometry of sine angle and cosine angle so you can see how sin (angle) = opposite/hypotenuse, which is hard to explain without imagery.
@@astropictionary3988 My lifetime's STUDENT-handicap is only slightly MORE advantaged than Nig-awanity ( all religiosities, self-governmental agencies incapatibilities' becoming unblinded compatible :: beyond nowaday's veil predestinated into non-gression, progressively foundationed all futuristics) . Thank you for gracious hope advancing my adjustments to avail potential's purposed for more of your remedial professorship. ).
Pool-hall village's jargon, we will LEAVE OURSELVES OUT FROM BEHIND THE TABLE'S 8-BALL.. ( -er )....... then, not whenever and not if ever!!! Thanks for participating FAITH within tangibility optional God's Family Angels. Love you amidst my predicament regardless, of impediments today.
actually, You have NOT explained how to measure the ANGLE "p"
Correct! How to measure is in the textbook. You would need to purchase it and i show how to measure.