Speed for maximum range
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Speed for maximum range
A co-worker has an interview coming up with Cathay Pacific. Since I had been through the interview process, I offered my assistance on a few topics, since I see myself as somewhat knowedgeable in technical fields.
The topic of conversation turned to what should be done to cruise speed as a heavy jet transport burns fuel and becomes lighter as the flight progresses. I was sure that I read that you climb to reduce fuel burn further, and reduce airspeed. Can someone confirm this is correct before I send this guy off with some bad info?
As proof, I offered to draw a thrust required vs. speed curve. I explained graphically how a line drawn from the origin (in no-wind conditions) to a point which is tangent to the thrust curve. The corresponding airspeed on the x-axis represents best range.
The part I cannot remember exactly is does the thrust-required curve move down and to the left as fuel is burned? This would produce the reduced airspeed which I am pretty sure is correct. We logically figured this is correct, since thrust required would be less at lighter weights.
Thanks
The topic of conversation turned to what should be done to cruise speed as a heavy jet transport burns fuel and becomes lighter as the flight progresses. I was sure that I read that you climb to reduce fuel burn further, and reduce airspeed. Can someone confirm this is correct before I send this guy off with some bad info?
As proof, I offered to draw a thrust required vs. speed curve. I explained graphically how a line drawn from the origin (in no-wind conditions) to a point which is tangent to the thrust curve. The corresponding airspeed on the x-axis represents best range.
The part I cannot remember exactly is does the thrust-required curve move down and to the left as fuel is burned? This would produce the reduced airspeed which I am pretty sure is correct. We logically figured this is correct, since thrust required would be less at lighter weights.
Thanks
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G'day Geardoor
I believe the way to achieve absolute maximum range for a given quantity of fuel is to continuously climb (obviously not always possible in a practical sense). The two considerations are the airframe and the engines.
Firstly for the airframe to be operated most efficiently it must be flown at the optimum AoA which will require the speed to be reduced as fuel is burnt.
Secondly the engines need to be operated at their 'design point' which is their most efficient regime. A continuous climb allows the engines to remain at the constant, most efficient RPM.
Beware: possibly completely untrue.
I believe the way to achieve absolute maximum range for a given quantity of fuel is to continuously climb (obviously not always possible in a practical sense). The two considerations are the airframe and the engines.
Firstly for the airframe to be operated most efficiently it must be flown at the optimum AoA which will require the speed to be reduced as fuel is burnt.
Secondly the engines need to be operated at their 'design point' which is their most efficient regime. A continuous climb allows the engines to remain at the constant, most efficient RPM.
Beware: possibly completely untrue.
Several factors are converging.
1. The airframe experiences the least drag ie is most efficient, at a certain AoA so flight must be maintained at this AoA
2. For given weight the aircraft needs an equal & opposite amount of lift. Increased weight --> increased lift. Can't do it by increasing AoA due (1) above.
3. This means that speed is the other variable that can be controlled to change lift. Go faster (CAS or EAS, can't remember. Logically I vote for EAS but am happy to be corrected) --> more lift, so heavier a/c can achieve the required amount of lift while staying at the most efficient AoA by going faster. In other words, for a given weight there will be a particular speed that corresponds to the most efficient AoA.
4. The lift required will result in an amount of drag being produced, proportional to the lift.
5. Thrust must be supplied to to counter the drag so more weight --> more lift --> more drag --> more thrust needed, however only *just* enough thrust must be supplied or the a/c won't maintain the speed that corresponds to the most efficient AoA *unless* the excess thrust is converted into altitude.
6. There is a limit on how high the airframe can go before high speed aerodynamic effects result in a loss of efficiency ie speed of sound effects.
At the same time engine efficiency has an effect.
If turbine:
7. A turbine is most efficient when operated near max. RPM. Trouble is that at low altitudes this produces more thrust than is required to counter drag (see #5 above) so the engine must be throttled to restrict its output with a resulting cost in efficiency.
8. For a given RPM a turbine loses output as altitude increases. At some point as altitude is gained its reduced outpoint will coincide with what the airframe needs to balance drag at the most efficient AoA. This is the optimum altitude .
9. As fuel is burnt the a/c obviously gets lighter, requiring less lift & less resultant thrust. The choice is to reduce AoA (bad, moves away from most efficient AoA) or reduce speed (good at keeping best AoA but with a catch: carries with it a reduced requirement for thrust.)
10. Second option: Reduce thrust & lose some engine efficiency? Or climb higher until thrust is reduce 'naturally'. Second option is the one that wins. Climbing also has a gain w.r.t increased TAS for a given IAS/CAS/EAS althought the reduced IAS may mask this effect.
End result: For the most efficient flight the aircraft will fly at a reducing speed along with a continuous climb profile.
edited to correct an error
1. The airframe experiences the least drag ie is most efficient, at a certain AoA so flight must be maintained at this AoA
2. For given weight the aircraft needs an equal & opposite amount of lift. Increased weight --> increased lift. Can't do it by increasing AoA due (1) above.
3. This means that speed is the other variable that can be controlled to change lift. Go faster (CAS or EAS, can't remember. Logically I vote for EAS but am happy to be corrected) --> more lift, so heavier a/c can achieve the required amount of lift while staying at the most efficient AoA by going faster. In other words, for a given weight there will be a particular speed that corresponds to the most efficient AoA.
4. The lift required will result in an amount of drag being produced, proportional to the lift.
5. Thrust must be supplied to to counter the drag so more weight --> more lift --> more drag --> more thrust needed, however only *just* enough thrust must be supplied or the a/c won't maintain the speed that corresponds to the most efficient AoA *unless* the excess thrust is converted into altitude.
6. There is a limit on how high the airframe can go before high speed aerodynamic effects result in a loss of efficiency ie speed of sound effects.
At the same time engine efficiency has an effect.
If turbine:
7. A turbine is most efficient when operated near max. RPM. Trouble is that at low altitudes this produces more thrust than is required to counter drag (see #5 above) so the engine must be throttled to restrict its output with a resulting cost in efficiency.
8. For a given RPM a turbine loses output as altitude increases. At some point as altitude is gained its reduced outpoint will coincide with what the airframe needs to balance drag at the most efficient AoA. This is the optimum altitude .
9. As fuel is burnt the a/c obviously gets lighter, requiring less lift & less resultant thrust. The choice is to reduce AoA (bad, moves away from most efficient AoA) or reduce speed (good at keeping best AoA but with a catch: carries with it a reduced requirement for thrust.)
10. Second option: Reduce thrust & lose some engine efficiency? Or climb higher until thrust is reduce 'naturally'. Second option is the one that wins. Climbing also has a gain w.r.t increased TAS for a given IAS/CAS/EAS althought the reduced IAS may mask this effect.
End result: For the most efficient flight the aircraft will fly at a reducing speed along with a continuous climb profile.
edited to correct an error
Last edited by Tinstaafl; 19th Feb 2005 at 15:20.
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Several major points to be considered in isolation, and then merged.
.1. The optimum speed for maximum range cruise (MRC) at ANY altitude is found at the tangent of a line drawn from the origin 0/0 to the Total Drag / Thrust Required Vs TAS Curve. As weight reduces, the total drag curve moves closer to the 0/0 origin in both axes, i.e. Downwards and to the Left.
At Lower altitudes where MRC speed is below Mcrit, Maximum Range EAS will be constant for a given weight, regardless of altitude, thus Maximum Range CAS for a given weight increases with altitude. As weight reduces, EAS, and thus CAS, reduce for a fixed or increasing altitude. AoA is the dominant factor. For a given weight, Thrust Required remains constant with increasing altitude.
At Higher altitudes where MRC speed is above Mcrit, the Drag / Thrust Required Vs TAS Curve point of tangency is in the region where Drag / Thrust Required rises more steeply due to wave drag, and is therefore at a lower EAS, and defined by Mach No. As altitude increases, MRC Mach No. INCREASES, the speed schedule is no longer constant. AoA is no longer the dominant factor, wave drag becomes much more prominent in assessing the MRC speed schedule as altitude increases above the level at which MRC EAS = Mcrit. For a given weight, Thrust Required increases with increasing altitude.This is the region where Jet aeroplanes spend most of their time.
.2. Where a Head or Tail Wind component is introduced, the Total Drag Curve moves laterally to the Left or Right, depending upon the wind component. A Tail Wind effectively moves the 0/0 origin to the Left, creating a lower point of tangency, and a LOWER MRC speed, whether it be defined as EAS or Mach No. A Head Wind effectively moves the 0/0 origin to the Right, creating a higher point of tangency and a HIGHER MRC speed, whether it be defined as EAS or Mach No.
Thus far, we have seen the relationship between Thrust required to produce the MRC speed schedule. Fuel Flow is required to produce Thrust.
.3. Jet engines have one particular speed (be it N1, N2, N3, or a combination of all of them) where Thrust Specific Fuel Consumption (TSFC) is at it's optimum, i.e. the engine speed where the required thrust is generated for the least Fuel Flow. This speed varies but is usually in the vicinity of 90-93%, NOT maximum. Above and below the optimum TSFC engine speed, it costs more in fuel to produce the requisite thrust. Maximum Range Cruise at low levels requires the same thrust as at somewhat higher levels, but at an engine speed somewhat below optimum TSFC speed, and therefore fuel-inefficient. As we climb higher, engine speed must increase to produce the same thrust for MRC, but at an engine speed approaching optimum TSFC speed. When MRC speed is achieved at an engine speed which offers the optimum TSFC, the aircraft is at OPTIMUM ALTITUDE. Any further climb above this level requires engine speed to increase above optimum TSFC R.P.M., and fuel consumed per mile increases again, just as it did at lower levels. Cruise above OPTIMUM LEVEL will often be at LOWER Fuel Flows, looks good, but it is the fuel consumed per mile that matters, not Fuel Flow in isolation, as would be the case for holding.
As mentioned earlier, when cruising at levels where MRC is defined by Mach No., for a given weight, Thrust Required increases with increasing altitude, but, up to the Optimum Level, TSFC efficiency improves at a greater rate than does drag rise.
That's the short answer, there's much more.
Good luck with CX,
Old Smokey
.1. The optimum speed for maximum range cruise (MRC) at ANY altitude is found at the tangent of a line drawn from the origin 0/0 to the Total Drag / Thrust Required Vs TAS Curve. As weight reduces, the total drag curve moves closer to the 0/0 origin in both axes, i.e. Downwards and to the Left.
At Lower altitudes where MRC speed is below Mcrit, Maximum Range EAS will be constant for a given weight, regardless of altitude, thus Maximum Range CAS for a given weight increases with altitude. As weight reduces, EAS, and thus CAS, reduce for a fixed or increasing altitude. AoA is the dominant factor. For a given weight, Thrust Required remains constant with increasing altitude.
At Higher altitudes where MRC speed is above Mcrit, the Drag / Thrust Required Vs TAS Curve point of tangency is in the region where Drag / Thrust Required rises more steeply due to wave drag, and is therefore at a lower EAS, and defined by Mach No. As altitude increases, MRC Mach No. INCREASES, the speed schedule is no longer constant. AoA is no longer the dominant factor, wave drag becomes much more prominent in assessing the MRC speed schedule as altitude increases above the level at which MRC EAS = Mcrit. For a given weight, Thrust Required increases with increasing altitude.This is the region where Jet aeroplanes spend most of their time.
.2. Where a Head or Tail Wind component is introduced, the Total Drag Curve moves laterally to the Left or Right, depending upon the wind component. A Tail Wind effectively moves the 0/0 origin to the Left, creating a lower point of tangency, and a LOWER MRC speed, whether it be defined as EAS or Mach No. A Head Wind effectively moves the 0/0 origin to the Right, creating a higher point of tangency and a HIGHER MRC speed, whether it be defined as EAS or Mach No.
Thus far, we have seen the relationship between Thrust required to produce the MRC speed schedule. Fuel Flow is required to produce Thrust.
.3. Jet engines have one particular speed (be it N1, N2, N3, or a combination of all of them) where Thrust Specific Fuel Consumption (TSFC) is at it's optimum, i.e. the engine speed where the required thrust is generated for the least Fuel Flow. This speed varies but is usually in the vicinity of 90-93%, NOT maximum. Above and below the optimum TSFC engine speed, it costs more in fuel to produce the requisite thrust. Maximum Range Cruise at low levels requires the same thrust as at somewhat higher levels, but at an engine speed somewhat below optimum TSFC speed, and therefore fuel-inefficient. As we climb higher, engine speed must increase to produce the same thrust for MRC, but at an engine speed approaching optimum TSFC speed. When MRC speed is achieved at an engine speed which offers the optimum TSFC, the aircraft is at OPTIMUM ALTITUDE. Any further climb above this level requires engine speed to increase above optimum TSFC R.P.M., and fuel consumed per mile increases again, just as it did at lower levels. Cruise above OPTIMUM LEVEL will often be at LOWER Fuel Flows, looks good, but it is the fuel consumed per mile that matters, not Fuel Flow in isolation, as would be the case for holding.
As mentioned earlier, when cruising at levels where MRC is defined by Mach No., for a given weight, Thrust Required increases with increasing altitude, but, up to the Optimum Level, TSFC efficiency improves at a greater rate than does drag rise.
That's the short answer, there's much more.
Good luck with CX,
Old Smokey
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The most simple approach to this question is to note that the trangent to the drag curve touches at Vmrc. This is typically in the region of 1.32 Vmd. For jet aircraft this is the maximum still air range speed Vmrc.
As weight gradually decreases due to fuel burn, the value of Vmd gradually decreases. This causes the value of Vmrc to decrease.
The figure 1.32 is not absolutely accurate and for any given aircraft type will depend on the shape of the drag curve (deep and narrow or shallow and wide). 1.32 Vmd is simply the "conventional wisdom" figure for JAR exams.
As weight gradually decreases due to fuel burn, the value of Vmd gradually decreases. This causes the value of Vmrc to decrease.
The figure 1.32 is not absolutely accurate and for any given aircraft type will depend on the shape of the drag curve (deep and narrow or shallow and wide). 1.32 Vmd is simply the "conventional wisdom" figure for JAR exams.
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I seem to recall from the dim and distant past when I did the CFS course in the RAF that the requirement for maximum range in a jet was
1/SFC x IAS/Drag x TAS/IAS
1/SFC (SFC being Specific Fuel Consumption) is an engine requirement and is met by operating the engine at its design RPM (or equivalent) which is usually around 93%. Ideally therefore the aircraft should be climbed to an altitude where this RPM gives a cruise speed that satisfies the other two requirements. As weight is reduced the aircraft is allowed to cruise climb while maintaining design RPM.
IAS/Drag gives the best ratio of speed to drag and is derived by drawing a line from the origin to a tangent on the total drag curve. It is usually about 1.3 times min drag speed.
TAS/IAS is the reason that jets are more efficient at altitude. Flying the best IAS/Drag speed at higher levels increases the TAS/IAS ratio.
So, to get the best range immediately after take off (in an ideal world) the aircraft should be climbed to an altitude where design RPM gives the speed for best IAS/Drag ratio, the altiude thereby also giving the best TAS/IAS ratio. As fuel is used and weight decreases, min drag speed also decreases and with it the speed for best IAS/Drag ratio. The secret then is to allow the aircraft to gently climb at constant TAS while maintaining design RPM. This has the following benefits-
Design RPM and therefore engine efficiency are maintained at the optimum.
The IAS/Drag ratio is maintained at its optimum, although the IAS will gradually reduce.
As the IAS reduces the TAS/IAS ratio improves.
All of the above assumes no penalty from compressibility, but I did say that it was in an ideal world.
1/SFC x IAS/Drag x TAS/IAS
1/SFC (SFC being Specific Fuel Consumption) is an engine requirement and is met by operating the engine at its design RPM (or equivalent) which is usually around 93%. Ideally therefore the aircraft should be climbed to an altitude where this RPM gives a cruise speed that satisfies the other two requirements. As weight is reduced the aircraft is allowed to cruise climb while maintaining design RPM.
IAS/Drag gives the best ratio of speed to drag and is derived by drawing a line from the origin to a tangent on the total drag curve. It is usually about 1.3 times min drag speed.
TAS/IAS is the reason that jets are more efficient at altitude. Flying the best IAS/Drag speed at higher levels increases the TAS/IAS ratio.
So, to get the best range immediately after take off (in an ideal world) the aircraft should be climbed to an altitude where design RPM gives the speed for best IAS/Drag ratio, the altiude thereby also giving the best TAS/IAS ratio. As fuel is used and weight decreases, min drag speed also decreases and with it the speed for best IAS/Drag ratio. The secret then is to allow the aircraft to gently climb at constant TAS while maintaining design RPM. This has the following benefits-
Design RPM and therefore engine efficiency are maintained at the optimum.
The IAS/Drag ratio is maintained at its optimum, although the IAS will gradually reduce.
As the IAS reduces the TAS/IAS ratio improves.
All of the above assumes no penalty from compressibility, but I did say that it was in an ideal world.
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Thanks for all the responses. The one thing I forgot about at the time was that you need to have the engines operating at their max effficiency rpm. HTBJ makes reference to that at the beginning.
I'll forward the responses to my bud.
I'll forward the responses to my bud.
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hello every one,
for max range h optimum increases continually.
concord(snif!) used a continual slow climb schedule in cruise flight .
the subsonic barges try to approximate this by a stepclimb procedure, whereas level flight is assumed until eg, 2000ft below hoptimum, then climb of 4000ft(2000ft above hoptimum) is requested & so on.
for max range h optimum increases continually.
concord(snif!) used a continual slow climb schedule in cruise flight .
the subsonic barges try to approximate this by a stepclimb procedure, whereas level flight is assumed until eg, 2000ft below hoptimum, then climb of 4000ft(2000ft above hoptimum) is requested & so on.
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Keith. Ref your point about the shape of the drag curve affecting still air best range speed and statement about 'conventional wisdom'. If you assume the drag curve is made up of an x squared curve and a 1/x squared curve the tangent to the drag curve is always at 1.32vmd, it's calculus.
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ALEX,
The problem is not in the calculus, but is in the conclusion that you are drawing from it.
Mathematics can be a very powerful tool, but we must be very careful in how we use it. Whenever we try to apply it to real world situations we inevitably impose a number of simplifications. These can have a marked effect on the accuracy of the results we produce.
Your assumption that total drag is proportional to V squared plus 1/V squared is based in part on the assumption that CDP is constant. This is not actually true but is simply a reasonable approximation unless we want really accurate results. Because of this fact, total drag is not exactly proportional to V squared plus 1/V squared. If making accurate predictions of aerodynamic effects were simply a matter of doing a few sums then there would be far fewer wind tunnels in the world!
You are trying to prove one hypthesis by basing your argument on another hypthesis. Unless you are prepared to prove each one in turn then you will prove nothing.
The problem is not in the calculus, but is in the conclusion that you are drawing from it.
Mathematics can be a very powerful tool, but we must be very careful in how we use it. Whenever we try to apply it to real world situations we inevitably impose a number of simplifications. These can have a marked effect on the accuracy of the results we produce.
Your assumption that total drag is proportional to V squared plus 1/V squared is based in part on the assumption that CDP is constant. This is not actually true but is simply a reasonable approximation unless we want really accurate results. Because of this fact, total drag is not exactly proportional to V squared plus 1/V squared. If making accurate predictions of aerodynamic effects were simply a matter of doing a few sums then there would be far fewer wind tunnels in the world!
You are trying to prove one hypthesis by basing your argument on another hypthesis. Unless you are prepared to prove each one in turn then you will prove nothing.
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I do apologise KEITH. Your comment about drag curves being either deep and narrow or shallow and wide made me think you were unaware of the maths underlying the theory.
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Thank you ALEX.
It is worth noting that the way in which the drag coefficients actually vary with changes in speed is one of the factors that determine whether the drag curve is shallow and wide or narrow and deep. That is why I used the term.
It is worth noting that the way in which the drag coefficients actually vary with changes in speed is one of the factors that determine whether the drag curve is shallow and wide or narrow and deep. That is why I used the term.
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Ah, I was afraid you meant that. The variation of CD with mach number has more to do with the rapid increase of drag at MCRIT, as described by Old Smokey above, than the general shape of the drag curve.
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Thank you Alex, for getting the thread back into line. I was getting nervous when the discussion centred around V squared and 1/V squared, perfectly good at lower altitudes, but not above Mcrit, and, to reitterate my own words from my earlier post, "This is the region where Jet aeroplanes spend most of their time". (Above Mcrit)
As for discussion relating to Vmd's incorporation into the various formulae, there are a great number of high flying "Business Jet" type aircraft where Mcrit is BELOW Vmd defined by the low speed polars, and Minimum Drag speed is then defined by Mach Number, now we have to consider Mmd (Mach Number for minimum drag, i.e. flying on the "back" side of the conventional drag curve defined by EAS, but nevertheless at minimum total drag as defined by EAS and Mach Number).
One more spanner to throw into the works.
Regards,
Old Smokey
As for discussion relating to Vmd's incorporation into the various formulae, there are a great number of high flying "Business Jet" type aircraft where Mcrit is BELOW Vmd defined by the low speed polars, and Minimum Drag speed is then defined by Mach Number, now we have to consider Mmd (Mach Number for minimum drag, i.e. flying on the "back" side of the conventional drag curve defined by EAS, but nevertheless at minimum total drag as defined by EAS and Mach Number).
One more spanner to throw into the works.
Regards,
Old Smokey
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ALEX,
My use of the term "narrow and deep or shallow and wide" was deliberate, but you do not appear to understand its significance.
Let's imagine an aircraft for which CDP is zero and CDI is more than zero (impossible I know, but let's just consider it) If we set aside shock induced drag, then VMD will be at infinity. Now draw a tangent to that curve and it will not touch at 1.32 VMD.
Now let's consider one with CDI zero and CDP more than zero. In this case VMD is zero and the tangent will touch very close to zero. Neither of these cases give a tangent touching at 1.32 VMD.
Now let's look at something more realistic.
Sketch a typical situation with DI and DP crossing below the minimum drag value, and a tangent touching the total drag curve at 1.32 VMD.
Now imagine that we have reduced both CDI and CDP such that VMD remains constant. The whole of the total drag curve will be lower, shallower and and wider. But a tangent will no longer touch the curve at 1.32 VMD. It will touch closer to VMD.
Do the same again and the curve again becomes lower, shallower and wider. The tangent is now even closer to VMD.
Now let's do the opposite. Increase both CDI and CDP such that VMD remains constant. The curve will move upwards and be deeper and narrower. But now the tangent touches at more than 1.32 VMD.
If we now add shock induced drag, the main effect will be to make the right hand side of the curves steeper. This will limit the maximum value of the best range speed. As Old Smokey has pointed out, this could well mean that best range is at a speed lower than the VMD predicted by the x squared + 1/x squared curve.
As I said earlier, 1.32 VMD is simply conventional wisdom. It is not absolutely accurate for all aircraft and it is not set in stone.
My use of the term "narrow and deep or shallow and wide" was deliberate, but you do not appear to understand its significance.
Let's imagine an aircraft for which CDP is zero and CDI is more than zero (impossible I know, but let's just consider it) If we set aside shock induced drag, then VMD will be at infinity. Now draw a tangent to that curve and it will not touch at 1.32 VMD.
Now let's consider one with CDI zero and CDP more than zero. In this case VMD is zero and the tangent will touch very close to zero. Neither of these cases give a tangent touching at 1.32 VMD.
Now let's look at something more realistic.
Sketch a typical situation with DI and DP crossing below the minimum drag value, and a tangent touching the total drag curve at 1.32 VMD.
Now imagine that we have reduced both CDI and CDP such that VMD remains constant. The whole of the total drag curve will be lower, shallower and and wider. But a tangent will no longer touch the curve at 1.32 VMD. It will touch closer to VMD.
Do the same again and the curve again becomes lower, shallower and wider. The tangent is now even closer to VMD.
Now let's do the opposite. Increase both CDI and CDP such that VMD remains constant. The curve will move upwards and be deeper and narrower. But now the tangent touches at more than 1.32 VMD.
If we now add shock induced drag, the main effect will be to make the right hand side of the curves steeper. This will limit the maximum value of the best range speed. As Old Smokey has pointed out, this could well mean that best range is at a speed lower than the VMD predicted by the x squared + 1/x squared curve.
As I said earlier, 1.32 VMD is simply conventional wisdom. It is not absolutely accurate for all aircraft and it is not set in stone.
Last edited by Keith.Williams.; 23rd Feb 2005 at 09:35.
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Holy smoke... Reading with interest the posts that both Keith and Alex are making only fills me with wonder about how I managed to pass all those exams and get into the right hand seat!!Well done Keith you made sense to me at the time in Bournemouth... although now the crossword in the times during the cruise can break me out in a sweat....
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1.32 ~= (3)**(1/4)
The figure derives from an incompressible flow analysis where it can be demonstrated for max range the parasite drag is equal to three times the induced drag. This contrasts the case where at minimum drag, the two contributions are equivalent.
For a constant total lift, C_L is proportional to (1/V**2), V_mr can be shown to be equal to 3**(1/4)*V_md.
But its only an approximation good perhaps for M<=0.3.
The figure derives from an incompressible flow analysis where it can be demonstrated for max range the parasite drag is equal to three times the induced drag. This contrasts the case where at minimum drag, the two contributions are equivalent.
For a constant total lift, C_L is proportional to (1/V**2), V_mr can be shown to be equal to 3**(1/4)*V_md.
But its only an approximation good perhaps for M<=0.3.