For the real engineers here

wrote:

I'm thinking of a clean glider, one that might weigh 1500 pounds and
has a glide angle of say 1 in 25. At 50 miles an hour, that would mean
in an hour's time it might descend two miles (of course scale it
reasonable numbers, I chose those for ease of calculation). That means
it's losing about 1500 * 5280 * 2, or about 16 million foot pounds of
energy an hour.

Now if I add an engine swinging an 8 foot diameter prop, maybe as a
pusher, the question is, how big an engine for cruise only? A
horsepower is 550 foot lbs a second, or about 2 million foot pounds
an hour. If all of that is correct, it suggests with a 50% efficient
prop a little 16 horsepower engine could pretty much keep this thing
at constant altitude.

It passes the reasonableness test as far as I can see. Any serious
disagreements?

For those of you who do things in metric units? I went to school a
long long time ago, and here in the US I can buy a little Briggs and
Stanton (spelling?) engine with a horsepower rating, not a kilowatt
one.

well, seems to be correct. still, let me add some annotations:

- i'd calculate directly using power instead of energy. the installed power
you need is simply weight*sink speed/efficiency; in a formula:
P = W*w/eta = m*g*v/(E*eta) with the glide ratio E = Lift/Drag, m the mass
and g the gravitational acceleration

- i prefer SI units, for the simple benefit tp be able to calculate without
conversion factors. this eliminates a quite likely source of mistakes (ask
NASA...). a few years ago, while working in the US, i failed to calculate
the mass of a simple sheet of aluminum (don't laugh!); i had several
numbers for the material's density, but none in the combination of units
for volume and mass that i needed; so i decided it was safer to go via SI
and convert the mass back to ounces...

- the conversion hp-kW is simple: 1 kW = 1.34 hp (= 1.36 german PS) or
roughly 4/3 hp


cheers
uli

On Jun 26, 4:09 am, Uli wrote:
wrote:
I'm thinking of a clean glider, one that might weigh 1500 pounds and
has a glide angle of say 1 in 25. At 50 miles an hour, that would mean
in an hour's time it might descend two miles (of course scale it
reasonable numbers, I chose those for ease of calculation). That means
it's losing about 1500 * 5280 * 2, or about 16 million foot pounds of
energy an hour.

Now if I add an engine swinging an 8 foot diameter prop, maybe as a
pusher, the question is, how big an engine for cruise only? A
horsepower is 550 foot lbs a second, or about 2 million foot pounds
an hour. If all of that is correct, it suggests with a 50% efficient
prop a little 16 horsepower engine could pretty much keep this thing
at constant altitude.

It passes the reasonableness test as far as I can see. Any serious
disagreements?

For those of you who do things in metric units? I went to school a
long long time ago, and here in the US I can buy a little Briggs and
Stanton (spelling?) engine with a horsepower rating, not a kilowatt
one.

well, seems to be correct. still, let me add some annotations:

- i'd calculate directly using power instead of energy. the installed power
you need is simply weight*sink speed/efficiency; in a formula:
P = W*w/eta = m*g*v/(E*eta) with the glide ratio E = Lift/Drag, m the mass
and g the gravitational acceleration

- i prefer SI units, for the simple benefit tp be able to calculate without
conversion factors. this eliminates a quite likely source of mistakes (ask
NASA...). a few years ago, while working in the US, i failed to calculate
the mass of a simple sheet of aluminum (don't laugh!); i had several
numbers for the material's density, but none in the combination of units
for volume and mass that i needed; so i decided it was safer to go via SI
and convert the mass back to ounces...

- the conversion hp-kW is simple: 1 kW = 1.34 hp (= 1.36 german PS) or
roughly 4/3 hp

cheers
uli
I wouldn't argue with you about using SI units in professional
communications -- I do that all of the time -- but in this case I
started out with English units and it was easy to stay within them.
Also, and importantly, the question I asked was more easily understood
by most pilots here, and the more useful answers came back in the same
units. First rule of communication -- speak the language the spoken to
are most likely to understand!

It would have been fun to give the airspeed in furlongs per fortnight,
or for the spectroscopically inclined, nm/sec.

I do appreciate your comments, thanks. Now let's give the thread back
to the little boys with their spray cans.

On Jun 26, 4:09 am, Uli wrote:
wrote:
I'm thinking of a clean glider, one that might weigh 1500 pounds and
has a glide angle of say 1 in 25. At 50 miles an hour, that would mean
in an hour's time it might descend two miles (of course scale it
reasonable numbers, I chose those for ease of calculation). That means
it's losing about 1500 * 5280 * 2, or about 16 million foot pounds of
energy an hour.

Now if I add an engine swinging an 8 foot diameter prop, maybe as a
pusher, the question is, how big an engine for cruise only? A
horsepower is 550 foot lbs a second, or about 2 million foot pounds
an hour. If all of that is correct, it suggests with a 50% efficient
prop a little 16 horsepower engine could pretty much keep this thing
at constant altitude.

It passes the reasonableness test as far as I can see. Any serious
disagreements?

For those of you who do things in metric units? I went to school a
long long time ago, and here in the US I can buy a little Briggs and
Stanton (spelling?) engine with a horsepower rating, not a kilowatt
one.

well, seems to be correct. still, let me add some annotations:

- i'd calculate directly using power instead of energy. the installed power
you need is simply weight*sink speed/efficiency; in a formula:
P = W*w/eta = m*g*v/(E*eta) with the glide ratio E = Lift/Drag, m the mass
and g the gravitational acceleration

- i prefer SI units, for the simple benefit tp be able to calculate without
conversion factors. this eliminates a quite likely source of mistakes (ask
NASA...). a few years ago, while working in the US, i failed to calculate
the mass of a simple sheet of aluminum (don't laugh!); i had several
numbers for the material's density, but none in the combination of units
for volume and mass that i needed; so i decided it was safer to go via SI
and convert the mass back to ounces...

- the conversion hp-kW is simple: 1 kW = 1.34 hp (= 1.36 german PS) or
roughly 4/3 hp

cheers
uli
I wouldn't argue with you about using SI units in professional
communications -- I do that all of the time -- but in this case I
started out with English units and it was easy to stay within them.
Also, and importantly, the question I asked was more easily understood
by most pilots here, and the more useful answers came back in the same
units. First rule of communication -- speak the language the spoken to
are most likely to understand!

It would have been fun to give the airspeed in furlongs per fortnight,
or for the spectroscopically inclined, nm/sec.

I do appreciate your comments, thanks. Now let's give the thread back
to the little boys with their spray cans.

I wouldn't argue with you about using SI units in professional
communications -- I do that all of the time -- but in this case I
started out with English units and it was easy to stay within them.
Also, and importantly, the question I asked was more easily understood
by most pilots here, and the more useful answers came back in the same
units. First rule of communication -- speak the language the spoken to
are most likely to understand!

my direct answer wasn't in another "language"; it was pretty straight, i
think. the formula i suggested is independent of units; so it can used also
if one calculates with imperial units.

you asked for an answer by "real engineers"; that's what you got.

i'm sorry if i confused anyone by giving additional information...