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james_davis
28th Sep 2003, 01:53
Can anyone help with this ATPL question?

How does VMCG change with increasing field elevation and temperature?

A. Increases, because at a lower density a larger IAS is necessary to generate the required rudder force.
B. Increases, because VMCG is related to V1 and VR and those speeds increase if the density decreases.
C. Decreases, because the engine thrust decreases.
D. Decreases, because VMCG is expressed in IAS and the IAS decreases with TAS constant and decreasing density.

BUDGIE
28th Sep 2003, 03:53
I would go with answer C.

Thrust reduces with increased alt/temp. Well it would certainly be the case for Vmca so don't see why it would change for Vmcg. Sure someone will correct me though:O

BUDGIE

TheHustler
28th Sep 2003, 04:42
If you read the course books for your course properly then you will find the answer there boyo! This is called ...cheating!

DB6
28th Sep 2003, 16:08
Hustler, no it's called learning (unless he's sitting in the exam in which case I imagine the invigilator might notice his laptop or hear the gentle tap-tapping of keys).
Sometimes the answer is not always evident from the books. Do you know the answer? ******ed if I do without hurting my head!

NineEighteen
28th Sep 2003, 17:21
Answer (c) is correct.

If conditions are hot, high and humid the air is less dense. This reduces thrust and means when you lose the critical engine on takeoff there is less yawing effect from asymmetric thrust than there would be in denser air.

Vmcg concerns the calibrated airspeed for minimum control on the ground (i.e. using primary aerodynamic controls only to correct the yawing tendency). We know that flying surfaces are more effective the faster air flows over them so, if the yaw is less pronounced, the required airspeed for counteracting it need not be so high.

As elevation and temperature increase (Hot High and Humid) Vmcg decreases.

0918

james_davis
29th Sep 2003, 04:22
Thanks to all for the help, much obliged.

pugzi
29th Sep 2003, 18:45
This is rather simple. You can do it in your head and bash through the ideas, or you can look in the CAP698 (which is issued in the exam) on page 64 and 65 and you see the second table from the bottom gives you EXACTLY what your after. That table is for minV1 (CMG). It tells you the minimum V1 possible, which is actually Vmcg, so this table is actaully Vmcg, and it even tells you it is.
Know the CAP manual, out of 184 questions in the current feedback, about 30% can be answered by looking at the manual. Thats of course if you had an instructor that taught you this and knows performance. Know the CAP manual it can save a lot of time and nasty exam confussing thoughts.
Hope this helps.

YYZ
29th Sep 2003, 19:49
I do not know the answer but if I did I would tell you, as this is a chat forum for like-minded people either wanting to discuss aviation & the like, or after information on how to get into/on in the aforementioned field.

Thehustler I hope you never have to ask for help because if I remember correctly this is the second time you have knocked someone for asking a question there struggling with? What’s the problem with getting help when you’re struggling to see the light? Seems like common sense too me.
:suspect:

dorosenco
30th Sep 2003, 23:43
I would say that Vmcg has to do with the effectiveness of your rudder to maintain directional control on takeoff. At high altitudes and temperatures the density of the air is low, that means that you need a higher TAS (same IAS !) to achieve the same directional effectiveness.
Also, at high altitudes the output power of your engine is lower which means that in the case of an engine failure the amount of yaw generated will be less so less rudder input is required to compensate.

Answer A is incorrect because at high altitudes versus low altitudes you fly at the same IAS for same amount of lift, only TAS changes with air density.

Answer B is incorrect because Vmcg will determine V1.

Answer D is incorrect because the IAS is not decreasing at constant TAS and decreasing air density (suppose unlimited engine power).

Correct answer (and only possible answer) is C.