2016 Heilbronner Neckarcup – Doubles

Mateusz Kowalczyk and Igor Zelenay were the defending champions but chose to participate with different partners. Kowalczyk partnered Tomasz Bednarek while Zelenay partnered Rameez Junaid. Both failed to defend their title, with Kowalczyk losing to Zelenay in the first round and Zelenay losing to Sander Arends and Tristan-Samuel Weissborn in the quarterfinals.

Doubles
2016 Heilbronner Neckarcup
Champion Sander Arends
Tristan-Samuel Weissborn
Runner-up Nikola Mektić
Antonio Šančić
Final score6–3, 6–4

Sander Arends and Tristan-Samuel Weissborn won the title after defeating Nikola Mektić and Antonio Šančić 6–3, 6–4 in the final.

Seeds

  1. Julio Peralta / Horacio Zeballos (Quarterfinals)
  2. Andrej Martin / Hans Podlipnik (Quarterfinals)
  3. Ken Skupski / Neal Skupski (First round)
  4. Rameez Junaid / Igor Zelenay (Quarterfinals)

Draw

Key

First Round Quarterfinals Semifinals Final
1 J Peralta
H Zeballos
6 7
WC A Beck
Y Maden
2 5 1 J Peralta
H Zeballos
64 6 [5]
N Mektić
A Šančić
6 3 [10] N Mektić
A Šančić
77 4 [10]
LL I Sabanov
M Sabanov
3 6 [5] N Mektić
A Šančić
6 6
3 K Skupski
N Skupski
4 3 A Molteni
MÁ Reyes-Varela
2 2
M Arévalo
F Škugor
6 6 M Arévalo
F Škugor
3 2
A Molteni
MÁ Reyes-Varela
6 3 [10] A Molteni
MÁ Reyes-Varela
6 6
R Harrison
M Venus
4 6 [6] N Mektić
A Šančić
3 4
F Moser
M Zverev
64 68 S Arends
T-S Weissborn
6 6
S Arends
T-S Weissborn
77 710 S Arends
T-S Weissborn
2 6 [11]
T Bednarek
M Kowalczyk
3 1 4 R Junaid
I Zelenay
6 4 [9]
4 R Junaid
I Zelenay
6 6 S Arends
T-S Weissborn
2 6 [10]
WC D Brown
O Otte
6 3 [10] WC D Brown
O Otte
6 3 [5]
WC K Krawietz
S Stadler
3 6 [3] WC D Brown
O Otte
6 6
G Kretschmer
A Satschko
6 1 [8] 2 A Martin
H Podlipnik
4 4
2 A Martin
H Podlipnik
1 6 [10]
gollark: Yes, I know. It doesn't depend on that.
gollark: You fractionally get 1 million and fractionally die.
gollark: Well, it's good if 1e6/n - (equivalent monetary cost of dying)/n > 0. Multiply both sides by n and it's trivial.
gollark: 1e6 = 1 million.
gollark: The expected value is 1e6/n - (equivalent monetary cost of dying)/n. So whether it is a good choice depends on whether (equivalent monetary cost of dying is greater than 1e6 euros, which is no.

References

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