I have read that best lift /drag ratio gives maximum range which makes perfect sense to me. I have also read Carson speed gives the best fuel consumption. which doesn’t make sense because this would mean Carson speed would give better range. Seems to be a contradiction. Can anybody explain what this means in a way that makes sense thanks.
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$\begingroup$ Consider telling us what "FF" is. If it is "fuel flow", did you mean to say "maximum" rather then "minimum"? $\endgroup$– quiet flyerOct 17, 2022 at 2:42
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4$\begingroup$ For everyone whose's never heard of Carson speed: cafe.foundation/v2/pdf_tech/MPG.engines/… and mooneyspace.com/topic/15053-carsons-speed $\endgroup$– Ralph J ♦Oct 17, 2022 at 3:11
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Carson speed does not give the best fuel consumption (depending of course on your definition of "best"). To cover a given distance with the least amount of fuel in a piston airplane requires you to fly at the best L/D speed (indicated airspeed). Carson argues in the paper linked by @RalphJ that most piston GA aircraft are usually operated at speeds much higher than best L/D, because the best L/D speed is actually relatively slow, as those airplanes have to be designed for other requirements as well, like takeoff and landing performance.
Carson's idea was to look at how much speed you can gain from using more fuel, and the relationship between how much extra fuel you need to use to achieve a certain increase in speed is a curve that has a minium at the Carson speed. That means, between best L/D and the Carson speed you will use increasingly more fuel, and the fuel flow per distance will increase (less economic in terms of fuel cost), but the gains in airspeed (more economic in terms of time spent?) will get increasingly better, up to the point of the Carson speed. Beyond that point you will have to increase your fuel flow more and more to gain more airspeed.
To make this a little clearer and put some numbers on it, that may or may not resemble a normal light aircraft.
- mass: 1000 kg
- wing area: 12 m²
- aspect ratio 6, efficiency 0.85, $c_D0$ 0.05
- fuel mass 75 kg, energy content 40 MJ/kg, propulsion efficiency 0.3
Maximum endurance:
- airspeed: 29 m/s
- endurance: 6.73 h, fuel flow: 11.1 kg/h
- range: 703 km, fuel flow: 10.7 kg/100 km
- "fuel flow per airspeed": 0.369 (kg m)/(100 km s)
Maximum range:
- airspeed: 39 m/s (34% faster than max. endurance)
- endurance: 5.85 h, fuel flow: 12.8 kg/h
- range: 821 km, fuel flow: 9.1 kg/100 km
- "fuel flow per airspeed": 0.233 (kg m)/(100 km s)
Carson's speed:
- airspeed: 51 m/s (31% faster than max. range)
- endurance: 3.86 h, fuel flow: 19.4 kg/h (52% more than max. range!)
- range: 709 km, fuel flow: 10.6 kg/100 km (16% more than max. range!)
- "fuel flow per airspeed": 0.208 (kg m)/(100 km s) (11% less than max. range!)
It's the same as if you're driving a car on the highway. Let's take one that e.g. uses 5 L / 100 km when driving 80 km/h. Your range with a 50 L tank would be 1000 km, but let's say you only want to go 500 km, which would take more than 6 hours. If you now choose to go 130 km/h, fuel consumption will go up to e.g. 6 L / 100 km, enabling you to cover the 500 km in a bit less than 4 hours, but the fuel consumption would go up - both per distance, as given above, and per time (6 L / 100 km at 130 km/h is 7,8 L/h, whereas 5 L / 100 km at 80 km/h is only 4 L/h - this shows the large difference in power required).
Your speed increases more than your fuel consumption per distance, that's why the higher speed can be considered "better" if you value your time more than your money. Typically want to cover a certain distance by getting from one place to the other, not by simply driving for around for a given amount of hours.
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$\begingroup$ Sorry still don’t understand. A site I was reading states that “Hence, at Carson speed we may consider the speed for which the ratio fuel flow/knot of airspeed reaches a minimum”. Doesn’t this mean a small increase in fuel consumption for a relatively greater increase in speed. Doesn’t that equate to a better fuel consumption and therefore a better range. $\endgroup$– RonaldOct 17, 2022 at 7:24
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1$\begingroup$ @Ronald An extreme example: up to Carson Speed, each additional knot costs you an additional 1 oz of fuel per hour and reduces your range slightly. Above Carson Speed, each additional knot costs you 1 gallon/hour and reduces your range drastically (like an afterburner - it's an after-prop!). The real world has curves, not dramatic inflection points like this, but that's the idea. The fact that additional airspeed becomes more costly above Carson Speed doesn't mean that it isn't somewhat costly (to FF & to range both) above the lower max range speed. $\endgroup$– Ralph J ♦Oct 17, 2022 at 15:12
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1$\begingroup$ @Ronald No, the best range is at best L/D. The idea is more "I want to fly a given distance from A to B, and my time is worth money.", so Carson is willing to spend more money on fuel while getting from A to B faster. Flying at Carson's speed as opposed to best L/D will decrease travel time by 25%, but only increase fuel consumption by 15%. And right around Carson's speed is where flying faster gives you the best "return on invest", meaning the increase in speed per additional fuel flow is greatest. $\endgroup$ Oct 18, 2022 at 2:55
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$\begingroup$ That site is wrong. The fuel flow (mass per time, proportional to power) will increase by more than 50%, but the total fuel used for flying a given distance will only increase by 16%, because you take ~30% less time to cover that distance. – Raketenolli $\endgroup$– RonaldOct 19, 2022 at 3:28
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$\begingroup$ Ok. This is my question. If total fuel used for a given distance will only increase by 16% for a 30 knot increase in airspeed why don’t I get a greater range at Carson speed than at best range speed. – Ronald $\endgroup$– RonaldOct 19, 2022 at 3:30
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There are three airspeeds that are related:
Ve –> Speed for maximum endurance -> maximum flight time per unit of fuel
Vr –> Speed for maximum range –> maximum distance per unit of fuel
Vc –> Carson's speed –> minimum fuel consumption per unit of airspeed
It can be demonstrated that the three airspeeds are linked by the factor $\sqrt[4]{3} \approx 1.316$
Thus:
$$ Vc \approx 1.316 \cdot Vr $$ $$ Vr \approx 1.316 \cdot Ve $$
Carson's original paper is 'Fuel efficiency of small aircraft' https://arc.aiaa.org/doi/abs/10.2514/3.57417?journalCode=ja
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$\begingroup$ Why would this speed now called "Carson speed" when for decades before it was known as optimum transport performance? The original at least says explicitly what it is. $\endgroup$ Oct 21, 2022 at 19:53