simple way to calculate climb gradient?

[Founder]

Does anyone have a smart simple way to calculate the climbgradient?

for example, will you clear an obstacle at 5000m that is 500 ft with a climbgradient of 5%? how do you calculate this?

Kind Regards
Tim

[chornedsnorkack]

Does anyone have a smart simple way to calculate the climbgradient?
for example, will you clear an obstacle at 5000m that is 500 ft with a climbgradient of 5%? how do you calculate this?
Kind Regards
Tim
500 feet divided by climb gradient of 5% is 10 000 feet. Less than 5000 m.

Finding the climb gradient itself, from simple physics viewpoint... you could take thrust/weight, substract drag/lift... and wouldn´t you thus arrive at the climb gradient?

Of course, presumably there are required safety margins scattered all around the computations!

[Founder]

500 feet divided by climb gradient of 5% is 10 000 feet. Less than 5000 m.
Finding the climb gradient itself, from simple physics viewpoint... you could take thrust/weight, substract drag/lift... and wouldn´t you thus arrive at the climb gradient?
Of course, presumably there are required safety margins scattered all around the computations!

So regarding my question:
with a climb gradient of 5% the aircraft will reach 500 ft after traveling 3200m, after traveling 5000m the aircraft should be at (500+274) 774 ft.?

160 m (500ft) / 5% = 3200m
5000-3200 = 1800m
x / 5% = 1800m
x = 90 m (274ft) (conversion m x 3,045 = ft)

correct?

[Old Smokey]

I think that you're putting the cart before the horse. First find the gradient to the obstacle, and then limit the aircraft weight for the environmental conditions to ensure that that gradient can be achieved.

For your example of a 500 foot obstacle at 5000 M, the obstacle-clear gradient is 500 / (5000 / .3048) X 100 = 3.05%

Now, ensure from your aircraft performance data that a gradient of 3.05% or more is available at the proposed Takeoff Weight. If not possible, limit the weight until the aircraft is capable of a Net Gradient of 3.05%

NOTE - 1st Segment considerations not made in this simplified response.

Regards,

Old Smokey

[Founder]

I think that you're putting the cart before the horse. First find the gradient to the obstacle, and then limit the aircraft weight for the environmental conditions to ensure that that gradient can be achieved.
For your example of a 500 foot obstacle at 5000 M, the obstacle-clear gradient is 500 / (5000 / .3048) X 100 = 3.05%
Now, ensure from your aircraft performance data that a gradient of 3.05% or more is available at the proposed Takeoff Weight. If not possible, limit the weight until the aircraft is capable of a Net Gradient of 3.05%
NOTE - 1st Segment considerations not made in this simplified response.
Regards,
Old Smokey

Your way makes more sense and by 1'st segment you're talking about adding the height of 35 ft to the obstacle clearance?

Kind Regards
Tim

[oldebloke]

SIMPLE, mulitply the Groundspeed by the required 'gradient' equals the rate of climb..EG.154knots(gross weight A320)x2.4%=400'rate o'climb..
or 2,4% 0f 6070(feet per Knot)x120knots=145'per nuatical mile x 120(2nautical miles /minute)=291'rate'o'climb..
Cheers:ok:

[oldebloke]

or in your case 5%x groundspeed(eg 120knots)=:ok: 600'rate o' climb..

[novicef]

In your calculation where did you get the .3048 from?

For your example of a 500 foot obstacle at 5000 M, the obstacle-clear gradient is 500 / (5000 / .3048) X 100 = 3.05%

I am just doing my ATPL's so its nice to know this stuff.

[AerocatS2A]

novicef, it is the conversion from metres to feet.

[novicef]

I would have thought mult by 3.28 would have been more obvious, instead of dividing by .3048 but then thats me.

[OzExpat]

0.3048 is the conversion factor to turn metres into feet. It is necessary to convert either the 500 feet obstacle height to metres, or the 5000 metres distance to feet. OS chose to convert the distance to the same units as the obstacle height.

[Old Smokey]

Founder,

You asked ' "Your way makes more sense and by 1'st segment you're talking about adding the height of 35 ft to the obstacle clearance?
"

The 35 ft is in-built into the system already. The 1st segment is significantly less gradient than the 2nd segment, due to the gear being down or in transit, i.e. the gear retraction phase. 1st segment will invariably "duck under" the Obstacle-clear plane from the runway end to the obstacle, so, what you need to do when considering the 1st segment (which you must) is find the height and distance at the end of the 1st segment, and compare THAT to the remaining distance and height to the obstacle to extract the gradient required.

Thanks OzExpat for clarifying the Metres to Feet conversion to obtain common units. My preference for using division by 0.3048 is that it is an exact conversion, whereas multiplication by the inverse is not, being a surd.

Ahhh John_T, I would have said that the 3rd segment gradient requirement was 0%, but the assumed Net Acceleration Height was DERIVED from the relationship between Net and Gross gradients (Nit picking mode today).

Regards,

Old Smokey

[novicef]

Your point taken. Your posts are very informative, thanks.

[OzExpat]

OS... my preference over the years has been to use a multiplier of 3.2808. How would this be less accurate than dividing by 0.3048? Surely just 6 in 1, half a dozen in the other?:confused:

[Tim Zukas]

A foot is defined as 0.3048 meters exactly (with an exception or two). So a meter isn't exactly 3.2808 feet.

[BOAC]

OK, Tim - I'll buy it - what is it, then? I'll amend my 'metric' PAs:)

[chasing767]

Founder !
Does anyone have a smart simple way to calculate the climbgradient?

for example, will you clear an obstacle at 5000m that is 500 ft with a climbgradient of 5%? how do you calculate this?

to make things easier...and leave all those feet conversion behind and answer your question directly :

5% gradient gives you 50m / 1 km . at 5000m distance your height will be 5 x 50m = 250m hence you will clear the obstacle by 100 m

[ChristiaanJ]

Re all those conversions....
At some point now lost in the mist of time, an inch was defined as being exactly 25.4 mm, with no further figures behind the decimal point.
Hence a foot is also exactly 0.3048 m.
The "inverse" conversion factors are not exact, so you're always doubtful about the precision of the result. That may not matter much in a discussion about altitude, but it can matter when dealing with aircraft components!