Trigonometry & Sunset - objective-c

Not sure whether this question is appropriate on this forum, but very sure there is someone who can answer this.
Reason I'm asking this, is that I failed to pay attention in school to anything dealing with sines and cosine stuff, and somehow I think this is the direction I have to look for.
So, I'm making this little clock app, in which I want to set the background color to more or less correspond with the amount of sunlight at a given time. It doesn't need to be highly accurate, so, I don't need the real time sun position. The simple reasoning is: At midnight, it is supposedly the darkest hour (black background, 0% white), and at noon, the sun is shining the brightest (white background, 100% white), and everything in between should set the background in an appropriate grey tint.
Input to the algorithm is just the hour, ranging from 0 to 24.
Thanks in advance.
PS, I'm writing the app in Objective-C, but I think the equation will be so simple, that that would be irrelevant.

I guess this is the wrong place for this question (as you as for mathematics, instead of programming).
But anyway, this is what you can do (as you seem to wish to work with a sine/cosine):
Do the hour *7,5
So 0->0, 12->90, 24->180
As the sin(0)=0, sine(90)=1 and sine(180) = 0, (values 0,90,180 are degrees, not radians). You will get a value ranging from 0-1 with a max at 12 o'clock and a minimum at 0h and 24h.
I do not know Objective-C, but i guess it will look something like this:
hourdegrees = 7.5 * Input(hour);
hourradians = M_PI * hourdegrees / 180;
density = sin(hour);
M_PI is the call-function for Pi(3.14...) in Objective C
Converting degrees to radians: know that 360degrees = 2*Pi radians (see here: Sine to radians Objective-C)
Hope you are something with this

Related

GPS distance: Pythagora's on an Equirectangular approximation vs Haversine fomula errors at different scales?

I'm trying to decide whether it makes cpu processing time sense to use the more complex Haversine formula instead of the faster Pythagorean's formula but while there seems to be a pretty unanimous answer on the lines of: "you can use Pythagora's formula for acceptable results on small distances but haversine is better", I can not find even a vague definition on what "small distances" mean.
This page, linked in the top answer to the very popular question Calculate distance between two latitude-longitude points? claims:
If performance is an issue and accuracy less important, for small distances Pythagoras’ theorem can be used on an equi­rectangular projec­tion:*
Accuracy is somewhat complex: along meridians there are no errors, otherwise they depend on distance, bearing, and latitude, but are small enough for many purposes*
the asterisc even says "Anyone care to quantify them?"
But this answer claims that the error is about 0.1% at 1000km (but it doesn't cite any reference, just personal observations) and that for 4km (even assuming the % doesn't shrink due to way smaller distance) it would mean under 4m of error which for public acces GPS is around the open-space best gps accuracy.
Now, I don't know what the average Joe thinks of when they say "small distances" but for me, 4km is definitely not a small distance (- I'm thinking more of tens of meters), so I would be grateful if someone can link or calculate a table of errors just like the one in this answer of Measuring accuracy of latitude and longitude? but I assume the errors would be higher near the poles so maybe choose 3 representative lattitudes (5*, 45* and 85*?) and calculate the error with respect to the decimal degree place.
Of course, I would also be happy with an answer that gives an exact meaning to "small distances".
Yes ... at 10 meters and up to 1km meters you're going to be very accurate using plain old Pythagoras Theorem. It's really ridiculous nobody talks about this, especially considering how much computational power you save.
Proof:
Take the top of the earth, since it will be a worst case, the top 90 miles longitude, so that it's a circle with the longitudinal lines intersecting in the middle.
Note above that as you zoom in to an area as small as 1km, just 50 miles from the poles, what originally looked like a trapezoid with curved top and bottom borders, essentially looks like a nearly perfect rectangle. In other words we can assume rectilinearity at 1km, and especially at a mere 10M.
Now, its true of course that the longitude degrees are much shorter near the poles than at the equator. For example any slack-jawed yokel can see that the rectangles made by the latitude and longitude lines grow taller, the aspect ratio increasing, as you get closer to the poles. In fact the relationship of the longitude distance is simply what it would be at the equator multiplied by the cosine of the latitude of anywhere along the path. ie. in the image above where "L" (longitude distance) and "l" (latitude distance) are both the same degrees it is:
LATcm = Latitude at *any* point along the path (because it's tiny compared to the earth)
L = l * cos(LATcm)
Thus, we can for 1km or less (even near the poles) calculate the distance very accurately using Pythagoras Theorem like so:
Where: latitude1, longitude1 = polar coordinates of the start point
and: latitude2, longitude2 = polar coordinates of the end point
distance = sqrt((latitude2-latitude1)^2 + ((longitude2-longitude1)*cos(latitude1))^2) * 111,139*60
Where 111,139*60 (above) is the number of meters within one degree at the equator,
because we have to convert the result from equator degrees to meters.
A neat thing about this is that GPS systems usually take measurements at about 10m or less, which means you can get very accurate over very large distances by summing up the results from this equation. As accurate as Haversine formula. The super-tiny errors don't magnify as you sum up the total because they are a percentage that remains the same as they are added up.
Reality is however that the Haversine formula (which is very accurate) isn't difficult, but relatively speaking Haversine will consume your processor at least 3 times more, and up to 31x more computational intensive according to this guy: https://blog.mapbox.com/fast-geodesic-approximations-with-cheap-ruler-106f229ad016.
For me this formula did come useful to me when I was using a system (Google sheets) that couldn't give me the significant digits that are necessary to do the haversine formula.

Calculating walking distance for user over time

I'm trying to track the distance a user has moved over time in my application using the GPS. I have the basic idea in place, so I store the previous location and when a new GPS location is sent I calculate the distance between them, and add that to the total distance. So far so good.
There are two big issues with this simple implementation:
Since the GPS is inacurate, when the user moves, the GPS points will not be a straight line but more of a "zig zag" pattern making it look like the user has moved longer than he actually have moved.
Also a accuracy problem. If the phone just lays on the table and polls GPS possitions, the answer is usually a couple of meters different every time, so you see the meters start accumulating even when the phone is laying still.
Both of these makes the tracking useless of coruse, since the number I'm providing is nowwhere near accurate enough.
But I guess that this problem is solvable since there are a lot of fitness trackers and similar out there that does track distance from GPS. I guess they do some kind of interpolation between the GPS values or something like that? I guess that won't be 100% accurate either, but probably good enough for my usage.
So what I'm after is basically a algorithm where I can put in my GPS positions, and get as good approximation of distance travelled as possible.
Note that I cannot presume that the user will follow roads, so I cannot use the Google Distance Matrix API or similar for this.
This is a common problem with the position data that is produced by GPS receivers. A typical consumer grade receiver that I have used has a position accuracy defined as a CEP of 2.5 metres. This means that for a stationary receiver in a "perfect" sky view environment over time 50% of the position fixes will lie within a circle with a radius of 2.5 metres. If you look at the position that the receiver reports it appears to wander at random around the true position sometimes moving a number of metres away from its true location. If you simply integrate the distance moved between samples then you will get a very large apparent distance travelled.for a stationary device.
A simple algorithm that I have used quite successfully for a vehicle odometer function is as follows
for(;;)
{
Stored_Position = Current_Position ;
do
{
Distance_Moved = Distance_Between( Current_Position, Stored_Position ) ;
} while ( Distance_Moved < MOVEMENT_THRESHOLD ) ;
Cumulative_Distance += Distance_Moved ;
}
The value of MOVEMENT_THRESHOLD will have an effect on the accuracy of the final result. If the value is too small then some of the random wandering performed by the stationary receiver will be included in the final result. If the value is too large then the path taken will be approximated to a series of straight lines each of which is as long as the threshold value. The extra distance travelled by the receiver as its path deviates from this straight line segment will be missed.
The accuracy of this approach, when compared with the vehicle odometer, was pretty good. How well it works with a pedestrian would have to be tested. The problem with people is that they can make much sharper turns than a vehicle resulting in larger errors from the straight line approximation. There is also the perennial problem with sky view obscuration and signal multipath caused by buildings, vehicles etc. that can induce positional errors of 10s of metres.

Calculating distance in m in xyz between GPS coordinates that are close together

I have a set of GPS Coordinates and I want to find the speed required for a UAV to travel between them. Doing this by calculating distance in x y z and then dividing by time to travel - m/s.
I know the great circle distance but I assume this will be incorrect since they are all relatively close together(within 10m)?
Is there an accurate way to do this?
For small distances you can use the haversine formula without a relevant loss of accuracy in comparison to Vincenty's f.e. Plus, it's designed to be accurate for very small distances. This can be read up here if you are interested.
You can do this by converting lat/long/alt into XYZ format for both points. Then, figure out the rotation angles to move one of those points (usually, the oldest point) so that it would be at lat=0 long=0 alt=0. Rotate the second position report (the newest point) by the same rotation angles. If you do it all correctly, you will find X equals the east offset, Y equals the north offset, and Z equals the up offset. You can use Pythagorean theorm with X and Y (north and east) offsets to determine the horizontal distance traveled. Normally, you just ignore the altitude differences and work with horizontal data only.
All of this assumes you are using accurate formulas to convert lat/lon/alt into XYZ. It also assumes you have enough precision in the lat/lon/alt values to be accurate. Approximations are not good if you want good results. Normally, you need about 6 decimal digits of precision in lat/lon values to compute positions down to the meter level of accuracy.
Keep in mind that this method doesn't work very well if you haven't moved far (greater than about 10 or 20 meters, more is better). There is enough noise in the GPS position reports that you are going to get jumpy velocity values that you will need to further filter to get good accuracy. The math approach isn't the problem here, it's the inherent noise in the GPS position reports. When you have good reports, you will get good velocity.
A GPS receiver doesn't normally use this approach to know velocity. It looks at the way doppler values change for each satellite and factor in current position to know what the velocity is. This works reasonably well when the vehicle is moving. It is a much faster way to detect changes in velocity (for instance, to release a position clamp). The normal user doesn't have access to the internal doppler values and the math gets very complicated, so it's not something you can do.

formulas in Projectile Motion?

how can i calculate angle to reach particular height?
suppose i want height 320.time is increasing as 0.1.
i am using h = (u sin(angle))^2 / 2g;
where can i put the time?
The inverse of the sin() function is called the arcsine, or sin-1 in mathematical notation. In many programming languages, it's available as asin().
From my answer to your previous question:
Where y is the height you want to reach (320 in this case), and assuming you're starting at y=0:
angle = arctan( 2*y / x )
where x is the distance on the X-AXIS between your starting point and the point where you want to reach that height, which is necessary in order to specify an angle.
If you really want me to, I can derive this one for you, but it follows directly from my answer to your previous question.
Also (since I can't comment answers yet I'm saying this here), you may be having issues getting an angle "less than 1" because you're trying to use degrees instead of radians. Many math libraries work in radians, so convert your angles.

How to calculate deceleration needed to reach a certain speed over a certain distance?

I've tried the typical physics equations for this but none of them really work because the equations deal with constant acceleration and mine will need to change to work correctly. Basically I have a car that can be going at a large range of speeds and needs to slow down and stop over a given distance and time as it reaches the end of its path.
So, I have:
V0, or the current speed
Vf, or the speed I want to reach (typically 0)
t, or the amount of time I want to take to reach the end of my path
d, or the distance I want to go as I change from V0 to Vf
I want to calculate
a, or the acceleration needed to go from V0 to Vf
The reason this becomes a programming-specific question is because a needs to be recalculated every single timestep as the car keeps stopping. So, V0 constantly is changed to be V0 from last timestep plus the a that was calculated last timestep. So essentially it will start stopping slowly then will eventually stop more abruptly, sort of like a car in real life.
EDITS:
All right, thanks for the great responses. A lot of what I needed was just some help thinking about this. Let me be more specific now that I've got some more ideas from you all:
I have a car c that is 64 pixels from its destination, so d=64. It is driving at 2 pixels per timestep, where a timestep is 1/60 of a second. I want to find the acceleration a that will bring it to a speed of 0.2 pixels per timestep by the time it has traveled d.
d = 64 //distance
V0 = 2 //initial velocity (in ppt)
Vf = 0.2 //final velocity (in ppt)
Also because this happens in a game loop, a variable delta is passed through to each action, which is the multiple of 1/60s that the last timestep took. In other words, if it took 1/60s, then delta is 1.0, if it took 1/30s, then delta is 0.5. Before acceleration is actually applied, it is multiplied by this delta value. Similarly, before the car moves again its velocity is multiplied by the delta value. This is pretty standard stuff, but it might be what is causing problems with my calculations.
Linear acceleration a for a distance d going from a starting speed Vi to a final speed Vf:
a = (Vf*Vf - Vi*Vi)/(2 * d)
EDIT:
After your edit, let me try and gauge what you need...
If you take this formula and insert your numbers, you get a constant acceleration of -0,0309375. Now, let's keep calling this result 'a'.
What you need between timestamps (frames?) is not actually the acceleration, but new location of the vehicle, right? So you use the following formula:
Sd = Vi * t + 0.5 * t * t * a
where Sd is the current distance from the start position at current frame/moment/sum_of_deltas, Vi is the starting speed, and t is the time since the start.
With this, your decceleration is constant, but even if it is linear, your speed will accomodate to your constraints.
If you want a non-linear decceleration, you could find some non-linear interpolation method, and interpolate not acceleration, but simply position between two points.
location = non_linear_function(time);
The four constraints you give are one too many for a linear system (one with constant acceleration), where any three of the variables would suffice to compute the acceleration and thereby determine the fourth variables. However, the system is way under-specified for a completely general nonlinear system -- there may be uncountably infinite ways to change acceleration over time while satisfying all the constraints as given. Can you perhaps specify better along what kind of curve acceleration should change over time?
Using 0 index to mean "at the start", 1 to mean "at the end", and D for Delta to mean "variation", given a linearly changing acceleration
a(t) = a0 + t * (a1-a0)/Dt
where a0 and a1 are the two parameters we want to compute to satisfy all the various constraints, I compute (if there's been no misstep, as I did it all by hand):
DV = Dt * (a0+a1)/2
Ds = Dt * (V0 + ((a1-a0)/6 + a0/2) * Dt)
Given DV, Dt and Ds are all given, this leaves 2 linear equations in the unknowns a0 and a1 so you can solve for these (but I'm leaving things in this form to make it easier to double check on my derivations!!!).
If you're applying the proper formulas at every step to compute changes in space and velocity, it should make no difference whether you compute a0 and a1 once and for all or recompute them at every step based on the remaining Dt, Ds and DV.
If you're trying to simulate a time-dependent acceleration in your equations, it just means that you should assume that. You have to integrate F = ma along with the acceleration equations, that's all. If acceleration isn't constant, you just have to solve a system of equations instead of just one.
So now it's really three vector equations that you have to integrate simultaneously: one for each component of displacement, velocity, and acceleration, or nine equations in total. The force as a function of time will be an input for your problem.
If you're assuming 1D motion you're down to three simultaneous equations. The ones for velocity and displacement are both pretty easy.
In real life, a car's stopping ability depends on the pressure on the brake pedal, any engine braking that's going on, surface conditions, and such: also, there's that "grab" at the end when the car really stops. Modeling that is complicated, and you're unlikely to find good answers on a programming website. Find some automotive engineers.
Aside from that, I don't know what you're asking for. Are you trying to determine a braking schedule? As in there's a certain amount of deceleration while coasting, and then applying the brake? In real driving, the time is not usually considered in these maneuvers, but rather the distance.
As far as I can tell, your problem is that you aren't asking for anything specific, which suggests that you really haven't figured out what you actually want. If you'd provide a sample use for this, we could probably help you. As it is, you've provided the bare bones of a problem that is either overdetermined or way underconstrained, and there's really nothing we can do with that.
if you need to go from 10m/s to 0m/s in 1m with linear acceleration you need 2 equations.
first find the time (t) it takes to stop.
v0 = initial velocity
vf = final velocity
x0 = initial displacement
xf = final displacement
a = constant linear acceleration
(xf-x0)=.5*(v0-vf)*t
t=2*(xf-x0)/(v0-vf)
t=2*(1m-0m)/(10m/s-0m/s)
t=.2seconds
next to calculate the linear acceleration between x0 & xf
(xf-x0)=(v0-vf)*t+.5*a*t^2
(1m-0m)=(10m/s-0m/s)*(.2s)+.5*a*((.2s)^2)
1m=(10m/s)*(.2s)+.5*a*(.04s^2)
1m=2m+a*(.02s^2)
-1m=a*(.02s^2)
a=-1m/(.02s^2)
a=-50m/s^2
in terms of gravity (g's)
a=(-50m/s^2)/(9.8m/s^2)
a=5.1g over the .2 seconds from 0m to 10m
Problem is either overconstrained or underconstrained (a is not constant? is there a maximum a?) or ambiguous.
Simplest formula would be a=(Vf-V0)/t
Edit: if time is not constrained, and distance s is constrained, and acceleration is constant, then the relevant formulae are s = (Vf+V0)/2 * t, t=(Vf-V0)/a which simplifies to a = (Vf2 - V02) / (2s).