Thursday's Pop quiz

[Mark]

Given a mean sea level designated as A,
if a distance of 1,173 ft is required for a
plane to take off when the temperature is
73 degrees F, and 1,356 ft when the temp.
is 86 degrees F, then what distance is required
when the temp. is 79.5 degrees, and given
a different elevation (B) in which a distance of
1,173 ft. is required for takeoff when the temp.
is 70 degrees F, and 1,356 ft.when the temp.
is 80 degrees, then at 75 degrees, what is
the difference in takeoff distance between
the two respective elevations?


---
Mark

[Mark]

On Nov 19, 2:27*pm, Mark > wrote:
> Given a mean sea level designated as A,
> if a distance of 1,173 ft is required for a
> plane to take off when the temperature is
> 73 degrees F, and 1,356 ft when the temp.
> is 86 degrees F, then what distance is required
> when the temp. is 79.5 degrees, and given
> a different elevation (B) in which a distance of
> 1,173 ft. is required for takeoff when the temp.
> is 70 degrees F, and 1,356 ft.when the temp.
> is 80 degrees, then at 75 degrees, what is
> the difference in takeoff distance between
> the two respective elevations?
>
> ---
> Mark

Since I didn't get any takers on my quiz,
then I'll go ahead and reveal that it was
sort of a trick question.

See, the elevation works out to give
us a take-off distance of...1,264.5 ft.

But when you do the math on the second
elevation, it also comes out to
1,264.5 ft., so,

There is no difference in distance. I
designed the problem to come out
with equal take-off distances.

So you say, then it's impossible, given
the temperature spread and density
altitudes for the performance
to vary like that.

Answer:

It was 2 different airplanes!

(not relevant to the problem)

---
Mark

[Mark]

On Nov 20, 7:25*pm, Mark > wrote:
> On Nov 19, 2:27*pm, Mark > wrote:
>
> > Given a mean sea level designated as A,
> > if a distance of 1,173 ft is required for a
> > plane to take off when the temperature is
> > 73 degrees F, and 1,356 ft when the temp.
> > is 86 degrees F, then what distance is required
> > when the temp. is 79.5 degrees, and given
> > a different elevation (B) in which a distance of
> > 1,173 ft. is required for takeoff when the temp.
> > is 70 degrees F, and 1,356 ft.when the temp.
> > is 80 degrees, then at 75 degrees, what is
> > the difference in takeoff distance between
> > the two respective elevations?
>
> > ---
> > Mark
>
> Since I didn't get any takers on my quiz,
> then I'll go ahead and reveal that it was
> sort of a trick question.
>
> See, the elevation works out to give
> us a take-off distance of...1,264.5 ft.
>
> But when you do the math on the second
> elevation, it also comes out to
> 1,264.5 ft., so,
>
> There is no difference in distance. I
> designed the problem to come out
> with equal take-off distances.
>
> So you say, then it's impossible, given
> the temperature spread and density
> altitudes for the performance
> to vary like that.
>
> Answer:
>
> *It was 2 different airplanes!
>
> (not relevant to the problem)
>
> ---
> Mark

Proved-

In the first scenerio of elevation A we establish
a relationship of the temperatures, such that the
unknown temperature/distance at 79.5 degrees
is expressed as 6.5/13, because the difference
between 73 degrees and 86 degrees is 13, and
the difference between 73 degrees and 79.5 is
6.5...therefore, 6.5 over 13.


Then you can do this in your head,
the difference between 1,173 ft. and 1,356 ft is
183.


The only thing missing now is the take-off
distance at 79.5. This is simply...X.


So, by interpolation, one can see-


6.5/13 = X/183


Then, 13X = 1189.5


So, X = 91.5


Therefore, 1,173 ft + 91.5 ft = 1264.5 ft at 79.5 degrees


That's the first answer. Now, thru reverse engineering,
we can use trickery and make another simple set of
data which yields the same answer, then apply it to
the landing strip scenerio.


Ok, we want an equation in which the answer comes
out to 91.5 ft again, and uses the same take-off
distance difference of 183.


So...


I establish a temperature relationship between 70
degrees and 80 degrees as...duh, 10. The unknown
distance temperature is now 75, which is simply
expressed as 5 over 10, or 5/10.


Now the numbers are rigged to come out again to
91.5 ft.


So...


5/10 = X/183


and 10X = 915


Therefore,


X = 91.5 again.


And like before, 1,173 + 91.5 = 1,264.5 ft. again.


Same answer for both take-off distances.

---
Mark