simple way to calculate climb gradient?
Thread Starter
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From: EU
simple way to calculate climb gradient?
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
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
Joined: Aug 2005
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From: Estonia
Originally Posted by 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
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
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!
Thread Starter
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From: EU
Originally Posted by chornedsnorkack
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!
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!
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?
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From: Australia
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
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
Thread Starter
Joined: Jan 2006
Posts: 364
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From: EU
Originally Posted by 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
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
Kind Regards
Tim
Joined: Apr 2004
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From: vancouver 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
or 2,4% 0f 6070(feet per Knot)x120knots=145'per nuatical mile x 120(2nautical miles /minute)=291'rate'o'climb..
Cheers
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From: Australia
Old Smokey.
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.
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.
PPRuNeaholic
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From: Cairns FNQ
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.
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From: Australia
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
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
Joined: Jan 2006
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From: Europe
Founder !
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
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?
for example, will you clear an obstacle at 5000m that is 500 ft with a climbgradient of 5%? how do you calculate this?
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
Joined: Jan 2005
Posts: 2,315
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From: France
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!
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!
