Hello. So then, the physical equation incorporating speed, distance and time(s): speed=distance/time(s) distance=speed*time(s) time(s)=speed/distance Running numbers from one to one hundred (times) discovers: 1/1 2/1,2/2 3/1,3/2,3/3... To produce a graph these values must be multiplied by one hundred i.e. (1/1)*100 (2/1)*100,(2/2)*100 (3/1)*100,(3/2)*100,(3/3)*100 Put simply the equation is: (((100-x)/100)*x) This may be confirmed on excel and produces a curvature: Enter Macros "for a=1 to 100 b=(((100-a)/100)*a) cells(a,1)=b next a" Is this curvature the Sun's passage across the sky? It should be possible to measure midday (when the Sun is at it's highest) by counting! When you get to one hundred you begin again from one. Does this show x to be measured at one hundred (counting one hundred from your present times and around produces the same number.) Just some information presented and some questions posed. Thanks in anticipation, Counter.

QUOTE="Counter, post: 3412421, member: 285114"]Hello. So then, the physical equation incorporating speed, distance and time(s): speed=distance/time(s) distance=speed*time(s) time(s)=speed/distance[/quote] Ok so far... Ok, so you're making a triangle of fractions... To make a graph of what? What is on the x axis of your graph and what is on the y axis? That's not an equation. An equation needs an equals (=) sign in it. An equation expresses an equality. What you have there is an expression. But what is it supposed to tell us? I'm not seeing the curvature. Please explain what you mean by "curvature". Are you plotting a graph, or what? Maybe. Tell me how your numbers relate to the Sun. I'm not getting what you mean. Please explain how I can tell it's midday by counting. Set down the step-by-step procedure I need to use, for example, or just explain it in clear words. What does the phrase "measured at 100" mean? Are you asking whether x (whatever it is) repeats cyclically or something like that? I don't understand what you're asking. No worries. Just clarify what you mean and it will be more likely that somebody can help you.

"Ok, so you're making a triangle of fractions..." Not simply a triangle: 1/1 2/1,2/2 3/1,3/2,3/3 ... 98/1,98/2,98/3,98/4,98/5,... 99/1,99/2,99/3,99/4,99/5,... 100/1,100/2,100.3,100/4,100/5,... "To make a graph of what? What is on the x axis of your graph and what is on the y axis?" 1-100 is on the x-axis. (((100-x)/100)*x) is on the y-axis. "I'm not seeing the curvature. Please explain what you mean by "curvature". Are you plotting a graph, or what?" You must enter macros, enter the above formula, then enter the spreadsheet and plot a graph. "Maybe. Tell me how your numbers relate to the Sun." At midday the Sun is at it's highest point. In the middle of the graph the curvature is at it's highest point. I was wondering whether the curve produced by the graph correlates to the Sun's passage across the sky. Perhaps instead the line is: Cells(1,a)=b "I'm not getting what you mean. Please explain how I can tell it's midday by counting. Set down the step-by-step procedure I need to use, for example, or just explain it in clear words." This is what I need to know. I need a closer: a finisher. I believe the counting procedure is cyclic, as you have written. When you get to one hundred you begin again from one. By counting one hundred (beginning again from one when you reach one hundred) you arrive at the same time. I am looking for a mathematical procedure that tells me when it's midday. Thanks for your reply.

To tell when it is midday - indeed to account for any physical phenomenon - you need observation. A mathematical procedure will get you nowhere without observation. What observation are you proposing to make?

Counter: Your graph is a parabola. What aspect of it do you think might correlate with the Sun's passage across the sky? For example, do you think you can somehow covert the y values to the angular position of the sun over the day, or something? Wouldn't it be more useful to scale the x axis to a 24 hour period, for a start?

Would it help to assume the Earth is a sphere (ie not flat) and the Sun goes round the centre of that sphere?

"Wouldn't it be more useful to scale the x axis to a 24 hour period, for a start?" Thank you for your reply JamesR. If you notice the twenty-sixth position minus the twenty-fourth position equals one. i.e. if the Sun is at that point it is one o'clock in the afternoon. And so one i.e. 26-24=1 27-23=2 28-24=3 ... 48-2=23 49-1=24 50-0=25 0-25 (i.e. 12-1)

Put a stick vertically in the ground. The shortest shadow happens at midday. Clocks are only approximates(so too, it would seem, mathematics?). Ancillary?: The earth ain't a sphere, it's a lumpy ovoid body that speeds up and slows down slightly over the course of a day. or I could be wrong?

This would be a very good way to generate the parabola. Lay a large piece of graph paper at the base of the sitck, plot the tip of the stick on the graph paper.

Maybe. Tell me what the current assumptions of your model of the sun are. So, what is this telling us about the Sun's passage across the sky, if anything?

Wolfram (bless) give us a graph of this function http://www.wolframalpha.com/input/?i=((100-a)/100)*a I will explain and support my proposed alternative shortly - after Counter has had a chance to look at this.

Looking at Counter's equation y=((100-a)/100)*a) where y is height and a is distance and assuming each unit is (say) one mile. The graph shows a body launched at 0, reaching maximum height at 50 miles along the x axis and landing after travelling 100 miles horizontally. Assuming midday is defined to be when the Sun is at it's highest then only people 50 miles from the launch point would get a proper midday. We may not know how big the Sun is but it certainly isn't small. The cannon required to launch it to a height of 25 miles is going to make one helluva bang - do we hear this bang? I suggest we don't. Also big bang when landing - not observed. The Sun lands 100 miles from launch point. Do we see the Sun being carried back ready for the next day? I don't think we do. Neither is there any evidence of any pit created by the Sun as it lands. Do we see new Suns being manufactured ready for the next day? I don't think so. To counter the above objections I propose one Sun that operates continuously on this path:- http://www.wolframalpha.com/input/?i=x^2+y^2=100000000

Impressive equation Confused2. The equation does not tell us anything about the path of the sun across the sky, rather the human mind counting can tell the time. One-hundred-and-one counted=+1 to the time. >--------> Counter.