Basic Maths question

sydthebeat

Hi all, wondering if you can give me an easier way, or a formulaic way to calculate this real world question that comes up a lot in my work.

Im looking for average room heights.

say i have a house that is 165 sq m

31.5 sq m @ 6.145 m high
15.7 sq m @ 3.945 m high
117.8 sq m @ 2.475 high

the way i currently do it is to find the average of the first two.
31.5 + 15.7 = 47.2 sq m
31.5/47.2 = 67%
(6.145 - 3.945)/100 * 67 = 1.474
average between the first two = 3.945 + 1.474 = 5.419

then get the average of this with the last height
47.2/165 = 28.6%
(5.419 - 2.475)/100 * 28.6 = 0.842
average = 2.475 + 0.842 = 3.317 m high


first question, is that correct?
and second question, is there an easier way or a formula that i could use?

thanks in advance

obriendj

that is one way to do it. adn you do have the right answer by my calculations

but I think there is a simpler way
especially if you have excel.


31.5 x 6.145 = 193.5675
15.7 x 3.945 = 61.9365
117.8 x 2.475 = 291.555
547.059/165 = 3.315509091

where 165 is the total of 31.5, 15.7 and 117.8

if you want the average sq metres just divide by (6.145 + 3.945 + 2.475)

sydthebeat

obriendj wrote: »

that is one way to do it. adn you do have the right answer by my calculations

but I think there is a simpler way
especially if you have excel.


31.5 x 6.145 = 193.5675
15.7 x 3.945 = 61.9365
117.8 x 2.475 = 291.555
547.059/165 = 3.315509091

where 165 is the total of 31.5, 15.7 and 117.8

if you want the average sq metres just divide by (6.145 + 3.945 + 2.475)

Thats exactly what i was looking for, thank you!!

i can write up a simple excel cal for that.

Yakuza

"Average" can mean different things to different people.
What has been calculated above would be the weighted average, but a no less meaningful "average" would simply be (6.145+3.945+2.475)/3 or 4.1883.

For what purpose do you use the average? That might point to which average you need.

From those ceiling heights, that sounds like a grand ole house (but I wouldn't want to have to heat it!)

sydthebeat

i need to fins the average ceiling heights. I carry out Building Energy rating assessments, so the average room height is an input value.
Yes its definitely the 'weighted' average thats required because the "average floor height" X "total perimeter" equals the dwelling volume, which is an important factor.
it wouldnt be a true reflection if i just divided the total by 3 and multiplied by the perimeter.

Yakuza

Fair enough, I was close to the mark when I said "I wouldn't want to heat it"

Just as an aside, average ceiling height of a house times the perimeter will not give its volume. It may be some metric used to work out the BER, but it's not the volume of the house.

Volume (of a cuboid space) is area x height, or in the case of the house referred to above, the sum of the volumes of the areas with different heights (their areas times their heights). Obriendj's method already computes the volume for you (545 m³ in this case).

As a layman in terms of building regs etc, I'd have thought the energy required to heat a house would be proportional to its volume, which would feed into the BER calc.