Descent Characteristics

Reynolds Number Regime

Using the chart generated for Drogue design and our current simulation results, a reasonable range of Altitudes and Mach numbers is 0 to 100,000 ft and 0.1 to 0.8, respectively. Using the 1976 standard atmospheric model with no corrections, this puts the Reynolds number (with a reference length of 0.1524 m corresponding to the 6in diameter of the falling tubes) between ~5000 and ~3,000,000. 

Relative Descent Dynamics

Booster Section

Assuming we are falling horizontally with flow conditions above the critical Reynolds number, we expect a drag coefficient (based on diameter) around 1.6. [1] This gives us a CdS of 0.53 m2, given a diameter of 6 in = 0.1524 m and a length of 85.73 in = 2.178 m (this neglects the fins). If we are falling axially, we expect a drag coefficient of around 0.8 [2]. Given a reference area of a circle with diameter 6 in (π*(0.0762m)2 = 0.0182 m2), we find that the CdS is 0.0146 m2.

It is difficult to predict in which configuration the booster will fall. On the one hand, if it falls horizontally, there will be greater drag-based restoring forces to counteract any perturbations to its configuration. On the other hand, if it falls axially, perturbations will have less of an effect due to smaller moment arms. This analysis is complicated by the fact that we are unsure of the rocket's orientation during deployment and the effect of altitude.

One way to predict things is to look at videos of drogue or tumble recovery experienced by other rockets. We have summarized the findings in the table below:

Video LinkFalling ConfigurationNotes
Tumble RecoveryMostly horizontally, at a bit of an angle 
Therion Flight Test 1Mostly horizontally, at a bit of an angle, tumbles around a lotUnder drogue with similar drag to body

Drogue

The drogue has a CdS of (this is at Mach 0.137) 0.662 m2.

Mission Package

Assuming that the mission package is falling nose down (it is planned to be stabilized that way via 2018 Reaction Wheel Payload System), it will have a coefficient of drag of approximately 0.2, given a fineness ratio of approximately 9 [2].Then, taking the same reference area as the axial configuration for the booster, we find a CdS of 0.00364 m2.

Descent Rates

Drogue

Given the previously mentioned CdS of 0.662 m2 at Mach 0.137 (which corresponds to approximately 150 ft/s at sea level), an air density of 1.225 kg/m3 at sea level, and a dry rocket mass of 89.83 lbs = 40.75kg (taken from the Hermes Mass Budget on 1/31/2018–this includes the drogue parachute itself, but we'll ignore this for now), we can calculate the approximate descent of the rocket under drogue when it reaches main deployment altitude (roughly sea level–I'm neglecting the 2000ft AGL and the altitude of Spaceport):

F_{D, dro} = \frac{1}{2} \rho C_{D} S V^2 = m_{dry}*g

F_{D, dro} = \frac{1}{2}*1.225 \frac{kg}{m^3}*0.662 m^2 *V^2 = 40.75 kg * 9.81 \frac{m}{s^2}

V 31.4 m/s = 103 ft/s

This velocity is well within an acceptable range per Requirement 3.1.1 of the DTEG.

One important caveat to this analysis is that this assumes the drogue will be pulling the entire rocket. As we see above, the CdS of the drogue and that of the booster section may be comparable, depending on the falling configuration. If this occurs, there will not be tension on the booster webbing, and the drogue therefore will not be "carrying its weight." Hermes' descent under drogue will occur faster. 1 To take this into account, we can bound the velocity neglecting the weight of the booster section. According to the mass budget, the propulsion system plus fin can weigh approximately 58.78 lb = 26.66 kg. Subtracting this from the previous weight we calculate:

F_{D, dro} = \frac{1}{2} \rho C_{D} S V^2 = m_{dry}*g

F_{D, dro} = \frac{1}{2}*1.225 \frac{kg}{m^3}*0.662 m^2 *V^2 = 14.09 kg * 9.81 \frac{m}{s^2}

V 18.46 m/s = 61 ft/s

This is a little lower than the suggested range, but not unreasonably so. Plus, as I mention in the footnote below, this is a pretty complex system and difficult to perfectly predict at all times. 

1 Admittedly, there is some super weird coupling in this system. If the drogue and mission package start to travel faster downwards than the booster section, eventually the webbing will be fully extended between the mission package and the booster and it will start to pull it down. Then maybe you'd enter some sort of periodic pattern of the webbing jerking the booster down and it falling behind again? I am not sure but I'm going to neglect this from my analysis.

Main (Adapted from PDR Fall 2017)

We will be using the Therion main parachute. Using test flight data from Project Raziel under this parachute, we determined the effective coefficient of drag of the rocket (which accounts for the drag on the body of the rocket, too.) 2. In the flight test we saw an approximate landing speed of 6.5 m/s, corresponding to a CD,eff,main of:3

F_{D, main} = \frac{1}{2} \rho C_{D, eff, main} S V^2 = m_{dry}*g

F_{D, main} = \frac{1}{2} 1.225 \frac{kg}{m^3} C_{D, eff, main} (0.96*(1.4478m)^2*\pi) (6.5 \frac{m}{s})^2 = (24 kg)*(9.81 \frac{m}{s^2})

CD,eff,main  1.44

Then, given its CdS of approximately 9, the main will definitely take all of the load of the dry rocket, so we can calculate for landing velocity:

F_{D, main} = \frac{1}{2}*1.225 \frac{kg}{m^3}* 1.44 * (0.96*(1.4478 m)^2 * \pi) *V^2 = 40.75 kg * 9.81 \frac{m}{s^2}

V 8.47 m/s = 28 ft/s

This is acceptable per Requirement 3.1.2 of the DTEG.

2 Because Raziel separated in two locations rather than just one, this number may vary from what will be seen on Hermes, which has a single separation point and a different height. We're using it here as an approximation for right now. One idea is to use a pitot tube to test for wind speed and rig the parachute up on a windy day with a fish scale or load cell. If a convenient opportunity during spring semester presents itself for this, we can do that. Or we can just use new flight test data from when we fly it (assuming it works...)

3 Within these calculations, reference area was chosen as the largest cross section of the parachute, which is approximated as having a 9.5 ft diameter (1.4478 m radius). We reduced this area by 4% to account for the approximate area lost to the spill hole. Mass was calculated using the after-burnout mass taken from the OpenRocket flight test model.

Relative Descent Rates

Drogue to Mission Package

Because the CdS of the drogue is 2 orders of magnitude greater than that on the mission package, we expect there to be a good deal of tension in the line connecting the drogue and the main. At sea level, we can calculate approximately how many pounds of force will be available to extract the deployment bag from the cup (assuming it hasn't fallen out prematurely) and then the main from the deployment bag. Velocity is chosen as the lower bound for steady state velocity

F_{extraction} = F_{D,dro} - F_{D, mp} = \frac{1}{2}*1.225 \frac{kg}{m^3}*(18.46 \frac{m}{s})^2*[CdS_{dro}-CdS_{main}]

F_{extr.} = 208.72\frac{kg}{ms^2}*[0.662 m^2-0.00364 m^2]

Fextr.  137.4 N = 31 lbs

Given the deployment bag from cup extraction force of roughly 4.52 lbs found during the December 2, 2017 Hermes Ground Test (this number may change due to a new deployment bag and different packing procedures) as well as the 2-3 pound parachute from deployment bag extraction force found during the Hermes Parachute Packing investigation, we expect that we will have more than sufficient extraction force.

Drogue to Booster

As previously mentioned, the variability in falling orientation makes it difficult to predict the relative ratio of CdS between the drogue and the booster. Given our current estimations, the drogue could see anywhere between 1.25x and 45x the amount of drag on the booster (if they are going the same velocity).

Resources

[1] DRAG OF CIRCULAR CYLINDERS FOR A WIDE RANGE OF REYNOLDS NUMBERS AND MACH NUMBERS

[2] Aerospaceweb.org, Drag of Cylinders and Cones

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