Derate and V-speeds

[stickN'rudder007]

Why is it that when the Derate method is used for take off , the resulting V-speeds are higher than those when maximum performance take off is used ? I would expect them to either stay the same or even decrease because the derate take off results in lower Vmcg.

[Empty Cruise]

Without derate, V2 is way below VL/D MAX. This is possible because the aircraft has so much power that flying at such a slow speed does not compromise performance.

When you derate, the performance edge is used to achieve a lower N1 & - most important - EGT during take-off. Now the aircraft needs a V2 that is a lot closer to VL/D MAX to meet the climb requirements. And since V2 needs to go up, up goes V1 and Vr as well.

Besides, I do remember something about the speed for max excess thrust increasing as thrust decreases - which just further adds to the requirement for increased speed.

Hope this helps - best regrads fm
Empty

[mutt]

If for exactly the same weights and conditions, you reduced the takeoff power by 10% (for example), some of the other variables have to change.

Calculating a Balanced Field Distance for the B777, 35C and 250,000 kgs, the 1 engine accelerate stop distance will increase from 8506 feet to 9413 feet.

Once this happens, the V-speeds will also change, the resulting V1 will change from 147.9 KIAS to 151.2 KIAS.

So in reality, you arent comparing apples with apples due to the change in the runway length required.

Mutt

[mutt]

Empty Cruise,

For the example i just gave, the V2 values are 160.5 (90B) and 159.9 (Derate), this is taken from the B777 AFM-DPI and is the opposite to what you just stated!

Mutt

[Empty Cruise]

Mutt,
Hmmmnn, might be you have so much punch in the 77 that you need a goodly derate for it to be limiting. Have you tried with the max allowable derate of 25% :confused:
Starts working straight away in a light twin (like the 73) :D

...or - maybe I'm not thinking straight :ouch: - how about "excess thrust to accelerate from V1 to V2 is less, therefore the V2/V1 ratio must shrink (if you still wanna reach V2 & tell the tale, that is)"?

Empty

[mutt]

Just did it.

25 % Derate
1 engine accel stop distance = 11109
V1 = 155.6
V2 = 159.4

30 % Derate
1 engine accel stop distance = 11910
V1 = 157.1
V2 = 159.5

So the V1 is increasing, the runway length required is increasing, the V2 is decreasing or reaches limit.

Just as a side issue, in the FAA world,the 25 % maximum derate value applies to the Assumed Temperature Derates and not a fixed Derate. Do the JAA apply the restriction to both?

Mutt

[Empty Cruise]

Yep, V2/V1 ratio it was, then :O

In the 73 QRP, I find three different tables for "Max allowed temp derate (25%)", one for 23,5K, one for 22K and one for 20K - so it's gotta be assumed temp only for that limit.

Would you know why the 25% limit came about? Changes in rudder forces when full thrust was selected after EFATO - or???

Empty

[oldebloke]

In the olde days with the Assumed Temp method,if one derated thrust to the 'assumed'temp one had to increase the V'speeds to the same 'performance levels.despite derate one would 'actually' reach V1 earlier due to actual cooler temp.So one had to increase the v1 speed to the 'assumed' to meet an even Stop/go...Blah Blah
Cheers:ok:

[ITCZ]

Without derate, V2 is way below VL/D MAX. This is possible because the aircraft has so much power that flying at such a slow speed does not compromise performance.
When you derate, the performance edge is used to achieve a lower N1 & - most important - EGT during take-off. Now the aircraft needs a V2 that is a lot closer to VL/D MAX to meet the climb requirements. And since V2 needs to go up, up goes V1 and Vr as well.
Besides, I do remember something about the speed for max excess thrust increasing as thrust decreases - which just further adds to the requirement for increased speed.
Hope this helps - best regrads fm
Empty

I'm not sure that those are the reasons, EC.

The aim of the V2 climb is to achieve a minimum gradient. The consideration is VL MAX or greatest coefficient of lift which occurs at greater AoA than VL/D MAX. A V2 climb is less than VL/D MAX because of this. If optimum rate of climb was the primary target, you could discuss VL/D MAX.

As to higher V1 in a derate/flex takeoff, remember that V1 must exceed Vmcg by a factor. If the aircraft is using less thrust, it will take longer to achieve Vmcg and require a longer field length in doing so, as mutt points out.

[john_tullamarine]

ITCZ,

"As to higher V1 in a derate/flex takeoff, remember that V1 must exceed Vmcg by a factor. If the aircraft is using less thrust, it will take longer to achieve Vmcg and require a longer field length in doing so"

You wouldn't like to expand on this at all, perhaps ?

[stickN'rudder007]

I think i got the reason based on Matt's explanation even though i could assume something slightly different :
Assuming you accelerate using Derate from brakes realease point till V1 (the original V1 calculated for max performance take off . Now this portion by it self will require more distance than when using max performance take off. Now if you continue from V1 your take off till reaching V2 ,again this portion by itself will be more than before becouse of the lower rate of acceleration .Your TODR overally increases significantly . If you were to stop following V1 then i would not expect the Deceleration distance it self to be much different than before .Your ASDR has icreased too ,but not as much as the TODR . So by definition since V1 is the speed at which it would take the same distance to either continue take off and reach screen height , or come to a full stop it would only make sence to increase V1 when using derate so as to equate the new distance from V1 to screen height with the decelaration distance from V1 to full stop .
I hope the way i wrote it makes sence ......

[Kristian17]

When choosing your derate temperature, do you therefore work out the maximum TO weight for the given rwy at max thrust, e.g. in Boeing 767 a max TO weight for a given rwy 150,000 kg, 15 degrees OAT and flaps 15, no slope (V1 138, Vr 142, V2 149). If your regulated TO weight is 130,000 kg do you derate your thrust until the V-speeds equal those in the Max TO conditions at full thrust? If this is correct, is the field length still balanced?

Thanks for your help
Kristian

[oldebloke]

K,Derate is simply using 'spare' runway to save the engines..
If your takeoff weight is anywhere near your limit forget it..If your 'actual weight'is light use the runway.Go into your charts at the actual weight check the temp alongside to see the 'saving' difference from Actual temp.In view of the improved 'braking'these days on the big boeings,one might not have to 'change the Vspeeds.Maybe the Analysis chart 'shows' the different(from actual,)speeds..As I said before,the 'up'ed V1 and V2 were to put the aircraft closer to the higher temp's position in the ACC/Stop runway case..:ok:

[mutt]

Oldebloke,

What model Boeings are you talking about???

In our case, we only deal with the Thrust Rating in use and the temperature to obtain the RTOW/V-speeds. We do not compare these values to ANY other conditions or speeds!

Kristian17,
If your aircraft weight =130,000, then you derate USING the assumed temperature method until the RTOW is equal to or greater than 130,000 kgs! V-speeds have nothing to do with it!


Mutt

[popay]

Kristian17, the purpose of derated take off is to generate higher take off weight if you are ASD limited in first place and additional benefit is that you still can perform a derated take off on contaminated RWY, which you cant do with FLEX. See the extraction from Airbus manual. Of course you still use derated option on the normal RWY to preserve engine life and so on.
Definition:
“AMJ 25-13 / AC 25-13
(4)(b) Derated takeoff thrust, for an aeroplane, is a takeoff thrust less than the maximum takeoff thrust, for which exists in the AFM a set of separate and independent takeoff limitations and performance data that complies with all requirements of Part 25.” In this case, “the thrust for takeoff is considered as a normal takeoff operating limit.”
Explanation:
A reduction in the minimum control speeds sometimes generates a takeoff performance benefit (higher MTOW) when taking-off on a short runway. Indeed, the decision speed V1 is the maximum speed at which it is still possible to reject the takeoff and stop the aircraft within the runway limits. Nevertheless, V1 must be greater than VMCG, and the Accelerate Stop Distance is often the most constraining limitation on a short runway. A reduction of the VMCG can then permit a reduction of the ASD for a given takeoff weight, and lead to better takeoff performance when the MTOW without derate is ASD/VMCG limited.
Procedure:
To carry out a derated takeoff, the actual takeoff weight and speeds have to be checked against the Maximum permissible takeoff weight computed for the given derate level (specific RTOW chart or equivalent computerized system).

One more thing is you don't select TOGA unless you are above F speed (Airbus) because the corresponding speeds are different from TOGA ones.
Cheers.:8

[Kristian17]

Thanks for all your help popay, mutt and oldebloke.

Kristian