How would I reverse this equation? - iphone

I’m using this equation to convert steps to estimated calories lost.
I now need to do the opposite and convert total calories to estimated steps.
This is the equation I’m using for steps to calories:
+(CGFloat) totalCalories:(NSUInteger)TotalStepsTaken weight:(CGFloat)PersonsWeight{
CGFloat TotalMinutesElapsed = (float)TotalStepsTaken / AverageStepsPerMinute; //Average Steps Per Minute is Equal to 100
CGFloat EstimatedCaloriesBurned = PersonsWeight * TotalMinutesElapsed * Walking3MphRate; //Walking3MphRate = 0.040;
return EstimatedCaloriesBurned;
}
It’s written in Objective-C, but I tried to make it as readable as possible.
All calculations are being done over a 1 hour period.
Thank you for you help.

Here's the algebra, I don't know Obj-C syntax...
Steps = round( Calories * StepsPerMinute / (Weight * WalkRate) );

Related

is there any formula to get travel time between two locations or from distance in flutter without using google maps

I have two locations and I have the formula to get the distance between them but I want to calculate the travel time between them without using google maps or any API calling. Is there any formula to do that in flutter?
I think the real travel time depends on a lot of factors, for example the road you take or the speed limit.
Whitout some complex elaboration, which are usually perfomed by a backend, I don't think it is possible to calculate it.
Anyway, since you have got the distance, you could simply estimate it by giving a supposed speed and then using basic physics formulas.
time = distance / speed.
You cannot find the travel distance between two locations without using the Google API.
However you can get the air distance in meters using this formula in flutter.
double calculateDistance(
double lat1, double lon1, double lat2, double lon2) {
var p = 0.017453292519943295;
var c = cos;
var a = 0.5 -
c((lat2 - lat1) * p) / 2 +
c(lat1 * p) * c(lat2 * p) * (1 - c((lon2 - lon1) * p)) / 2;
var temp = 12742 * asin(sqrt(a));
return (temp * 1000);
}

2D Motion vectors - calculate location of object at a given time

I am having problems understanding 2D motion vectors when moving certain objects at a given time. My knowledge of linear algebra is limited and I really don't know the exact search terms to look for, so I wanted to know whether anybody could help me or at least hint me in the right direction.
My problem looks like this:
I have two points, a startPoint, and an endPoint in space. They have each a specific location, denoted as (x_1, x_2) and (y_1, y_2) respectively. Both of these points have a time attached to it, named t_startPoint or t_endPoint, respectively. I now want to find out, for a given currentTime (= basically any point in time that is in between t_startPoint and t_endPoint), where exactly would a new point N be positioned on the connection line between those two points. I know the description is not trivial and that’s why I also added an image describing what I would like to do:
So far, this is what I have as my algorithm:
func update(_ time: Int64) {
let t_startPoint: Int64 = 1
let position_startPoint: = (1.0, 1.0)
let t_endPoint: Int64 = 5
let position_endPoint: Vector = (4.0, 5.0)
let currentTime = 3
let duration = t_endPoint - t_startPoint
let x = position_startPoint.x + ((position_endPoint.x - position_startPoint.x) / Float(duration)) * (Float(currentTime - t_startPoint))
let y = position_startPoint.y + ((position_endPoint.y - position_startPoint.y) / Float(duration)) * (Float(currentTime - t_startPoint))
//
However, no matter what I do, my objects keep overshooting, erratically moving back and forth, and I don't know where to start. Any help would be greatly appreciated!
For constant velocity moving there is relation:
(t-t1) / (t2-t1) = (x-x1) / (x2-x1)
x = x1 + (x2-t1) * (t-t1) / (t2-t1)
so your expresiion looks right. Check:
1 + (4-1) * (3-1) / (5-1) = 1 + 3 * 2 / 4 = 2.5 - exact middle, OK

Loss of precision when dividing doubles in swift

I try to find the steps between a min and a max value with a given step-size, using swift 2.1.
So we have a min and a max value, both of type Double. The step-size is a Double too. If min is 0.0 and max 0.5 with steps of 0.1, the result is 6, obviously.
But if I start with -0.1 as the minimum value, the result is 6 too. But should be 7, agree?
Here is my Playground example:
let min:Double = -0.1
let max:Double = 0.5
let step:Double = 0.1
var steps: Int {
return Int((max - min) / step) + 1
}
print("steps: \(steps)") // returns "steps: 6", but should be 7
The result is 6.99999999 if we use a Double for the steps variable. But this loss of precision only occurs when our min value is negative.
Do you know a workaround? I just don't want to round() each time I calculate with Doubles.
When you use Int() it forces truncation of your number, which always rounds towards zero. So, 6.9999 becomes 6 rather than 7, because it's closer to zero. If you use round() first it should help:
var steps: Int {
return Int(round((max - min) / step) + 1.0)
}
That's always not a good idea to calculate integral steps based on floating point ranges, you'll always encounter issues, and you won't be able to do much.
Instead I recommend to build your logic on integral steps, and calculate double values based on integral values (not vice versa as you do). I.e. you don't calculate integral step based on range, but you set your integral number of steps and calculate your double step.

Distance between two coordinates in php using haversine

I've looked around and seen mention of the haversine formula to determine distance between two coordinates (lat1, lng1) and (lat2, lng2).
I've implemented this code:
function haversineGreatCircleDistance(
$latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
// convert from degrees to radians
$latFrom = deg2rad($latitudeFrom);
$lonFrom = deg2rad($longitudeFrom);
$latTo = deg2rad($latitudeTo);
$lonTo = deg2rad($longitudeTo);
$latDelta = $latTo - $latFrom;
$lonDelta = $lonTo - $lonFrom;
$angle = 2 * asin(sqrt(pow(sin($latDelta / 2), 2) +
cos($latFrom) * cos($latTo) * pow(sin($lonDelta / 2), 2)));
return $angle * $earthRadius;
}
And am trying to determine:
1) what units this is returning? (goal being in feet)
2) is this equation written the right way?
For example what should be the distance between these two points?
(32.8940695525,-96.7926336453) and (33.0642604502, -96.8064332754)?
I'm getting 18968.0903312 from the formula above.
Thanks!
1) what units this is returning? (goal being in feet)
Whatever units in which you supply the Earth's radius.
2) is this equation written the right way?
Test it. You can compare your results with an existing Haversine formula implementation, like this one.

iPhone - AVAudioPlayer - convert decibel level into percent

I like to update an existing iPhone application which is using AudioQueue for playing audio files. The levels (peakPowerForChannel, averagePowerForChannel) were linear form 0.0f to 1.0f.
Now I like to use the simpler class AVAudioPlayer which works fine, the only issue is that the levels which are now in decibel, not linear from -120.0f to 0.0f.
Has anyone a formula to convert it back to the linear values between 0.0f and 1.0f?
Thanks
Tom
Several Apple examples use the following formula to convert the decibels into a linear range (from 0.0 to 1.0):
double percentage = pow (10, (0.05 * power));
where power is the value you get from one of the various level meter methods or functions, such as AVAudioPlayer's averagePowerForChannel:
Math behind the Linear and Logarithmic value conversion:
1. Linear to Decibel (logarithmic):
decibelValue = 20.0f * log10(linearValue)
Note: log is base 10
Suppose the linear value in the form of percentage range from [ 0 (min vol) to 100 (max vol)] then the decibelValue for half of the volume (50%) is
decibelValue = 20.0f * log10(50.0f/100.0f) = -6 dB
Full volume:
decibelValue = 20.0f * log10(100.0f/100.0f) = 0 dB
Complete mute:
decibelValue = 20.0f * log10(0/100.0f) = -infinity
2. Decibel(logarithmic) to Linear:
LinearValue = pow(10.0f, decibelValue/20.0f)
Apple uses a lookup table in their SpeakHere sample that converts from dB to a linear value displayed on a level meter.
I moulded their calculation in a small routine; see here.