# Thread: Help with Crafting an Equation

1. ## Help with Crafting an Equation

I have an internship measuring and analyzing telemetry data for the US military, and I've run into a phenomenon that I know can be represented as a mathematical equation. My current mathematical learning is not equipped to express it, however, so I would like some help.

First, we have an observed bearing, for example 094 relative to own instruments. Then we have an initial observable rate of change, or bearing rate, which for this example we'll use +0.5 degrees per minute. In addition we have an observable rate of change for the bearing rate itself, in this example we'll use +0.01 degrees per minute per minute. So at time 0 we have a bearing measurement of 094. At time +1 we have a measurement of 094.5. At time +2 we have a measurement of 095.01. At time +3 we have a measurement of 95.53 and so on. So if I want to calculate what bearing will measure at time +10 I take the initial bearing, add the bearing rate multiplied by ten then add the bearing rate change to that as it increases over time, in this case 0.01+0.02+0.03+0.04+0.05+0.06+0.07+0.08+0.09+0.1.

How can this be expressed as an equation? So far I've got something like y=a+bx+c... something, where y represents the final calculated measurement, a represents the initial bearing, b represents the initial bearing rate, x represents the time of the final calculated measurement and c(something) would represent the gradually increasing mathematical property of bearing rate rate. I'm guessing this is a function or an algorithm, but my current mathematical training is insufficient to put those into an equation. Any help would be much appreciated.

2. ## Re: Help with Crafting an Equation

Originally Posted by Mathophile
So at time 0 we have a bearing measurement of 094. At time +1 we have a measurement of 094.5. At time +2 we have a measurement of 095.01. At time +3 we have a measurement of 95.53 and so on.
It looks like you pretend that the bearing rate is increased in a stepwise manner: during the whole first minute the bearing rate is 0.5, during the whole second minute it is 0.51, during the third it is 0.52, etc. On the other hand, I would expect that the rate changes continuously from 0.5 to 0.51 during the first minute, from 0.51 to 0.52 during the second minute, etc. In may be easier to understand stepwise increments, but it does not make it easier to compute, probably the contrary. In particular, with stepwise changes you can't calculate the rate when the time is not a whole minute.

Originally Posted by Mathophile
So if I want to calculate what bearing will measure at time +10 I take the initial bearing, add the bearing rate multiplied by ten then add the bearing rate change to that as it increases over time, in this case 0.01+0.02+0.03+0.04+0.05+0.06+0.07+0.08+0.09+0.1.
Here you have 10 terms due to the bearing rate change (from 0.01 to 0.1). However, if you assume stepwise increments, there should be 9 terms with the last one being 0.09. Indeed, in your calculation in the previous quote, for time +1 the rate change did not make a contribution, for time +2 the contribution was 0.01, for time +3 it was 0.01 + 0.02.

If we assume that the bearing rate, which I denoted $\omega$, changes continuously, let's see how it depends on time.

Initially it is $b$ and then it grows linearly with $t$, so at time $t$ it becomes $b+ct$. The actual bearing is the shaded area under the graph (e.g., when $c = 0$, it is just $bt$). The triangle area is $ct^2/2$ (half of a rectangle with sides $t$ and $ct$), so the shaded area is $bt + ct^2/2$. Therefore, the bearing at time $t$ is $a+bt+ct^2/2$.

If you insist on stepwise increment to the bearing rate, then you can use the formula for the sum of arithmetic progression: $c+2c+\dots+ (t-1)c=c(1+2+\dots+t-1)=c(t-1)t/2$ when $t$ is an integer and $t>0$. Then the bearing at time $t$ is $a+bt+c(t-1)t/2$.

3. ## Re: Help with Crafting an Equation

Originally Posted by emakarov
It looks like you pretend that the bearing rate is increased in a stepwise manner: during the whole first minute the bearing rate is 0.5, during the whole second minute it is 0.51, during the third it is 0.52, etc. On the other hand, I would expect that the rate changes continuously from 0.5 to 0.51 during the first minute, from 0.51 to 0.52 during the second minute, etc.
Thank you! That's precisely what I was doing. I was assuming/pretending that bearing rate increasing in quanta because we only take measurements every minute rather than taking continuous bearings. I can see how assuming a constant increase would simplify the calculation.

Originally Posted by emakarov
In may be easier to understand stepwise increments, but it does not make it easier to compute, probably the contrary. In particular, with stepwise changes you can't calculate the rate when the time is not a whole minute.
Excellent point! I didn't think about that.

Originally Posted by emakarov
Here you have 10 terms due to the bearing rate change (from 0.01 to 0.1). However, if you assume stepwise increments, there should be 9 terms with the last one being 0.09. Indeed, in your calculation in the previous quote, for time +1 the rate change did not make a contribution, for time +2 the contribution was 0.01, for time +3 it was 0.01 + 0.02.

If we assume that the bearing rate, which I denoted $\omega$, changes continuously, let's see how it depends on time.

Initially it is $b$ and then it grows linearly with $t$, so at time $t$ it becomes $b+ct$. The actual bearing is the shaded area under the graph (e.g., when $c = 0$, it is just $bt$). The triangle area is $ct^2/2$ (half of a rectangle with sides $t$ and $ct$), so the shaded area is $bt + ct^2/2$. Therefore, the bearing at time $t$ is $a+bt+ct^2/2$.

If you insist on stepwise increment to the bearing rate, then you can use the formula for the sum of arithmetic progression: $c+2c+\dots+ (t-1)c=c(1+2+\dots+t-1)=c(t-1)t/2$ when $t$ is an integer and $t>0$. Then the bearing at time $t$ is $a+bt+c(t-1)t/2$.
Thank you! So, to recap, the continuous equation is: $a+bt+ct^2/2$; and the stepwise equation is: $a+bt+c(t-1)t/2$?

4. ## Re: Help with Crafting an Equation

Originally Posted by Mathophile
Thank you! So, to recap, the continuous equation is: $a+bt+ct^2/2$; and the stepwise equation is: $a+bt+c(t-1)t/2$?
Yes.

5. ## Re: Help with Crafting an Equation

Thanks a bunch! I was able to put that into an Excel spreadsheet to facilitate TMA. I very much appreciate it!